    # Viết chương trình in tất cả các hợp số giữa a và b trong C++

Giả sử chúng ta có một số n. Ta phải tìm hai số nguyên hỗn hợp (không nguyên tố) a và b sao cho a - b = n

Vì vậy, nếu đầu vào giống như n = 512, thì đầu ra sẽ là 4608 và 4096

## bước

Để giải quyết vấn đề này, chúng tôi sẽ làm theo các bước sau -

`print 10*n and 9*n.`

## Thí dụ

Chúng ta hãy xem triển khai sau đây để hiểu rõ hơn -

```#include
using namespace std;
void solve(int n){
cout<<10*n<<", "<<9*n;
}
int main(){
int n = 512;
solve(n);
}```

`512`

## đầu ra

`5120, 4608`

#includeusing tên không gian std;

## Sử dụng vòng lặp while

Trong ví dụ sau, chúng ta sẽ kiểm tra xem số 12 có phải là Hợp số hay không bằng cách sử dụng

### Thí dụ

#includeusing tên không gian std;

## Sử dụng vòng lặp do while

Trong ví dụ sau, chúng ta sẽ kiểm tra xem số 12 có phải là Hợp số hay không bằng cách sử dụng

### Thí dụ

#includeusing tên không gian std;

## Hợp Số Giữa Khoảng Đã Cho

Trong ví dụ sau, chúng ta sẽ tìm tất cả các Hợp số từ 1 đến 10

### Thí dụ

#includeusing namespace std; int main() { int start = 1; int end = 10; int count = 0; int i = 1; cout << "Composite Numbers between " << start <<" and " << end <<": \n"; for(start=start; start<=end; start++) { for(i=1; i<=start; i++) { if(start % i == 0) count++; } if(count > 2) cout << start << " "; count = 0; } return 0; }

### đầu ra

Các số hỗn hợp từ 1 đến 10. 4 6 8 9 10

## Kiểm tra xem số đã cho là số nguyên tố hay hợp số

Trong ví dụ sau, chúng ta sẽ kiểm tra xem số đã cho là số nguyên tố hay hợp số

### Thí dụ

#includeusing tên không gian std; . "; cin >> num; for(i=1; i<=num; i++) { if(num % i == 0) count++; } if(count == 2) cout << num <<" là số nguyên tố

### đầu ra

Nhập một số (int). 18 18 là hợp số

## Lời nhắc nhở

Xin chào các Nhà phát triển, chúng tôi đã bao quát gần 90% hàm Chuỗi và Câu hỏi phỏng vấn trên C++ với các ví dụ để học nhanh chóng và dễ dàng

Chúng tôi đang làm việc để bao hàm mọi Khái niệm Đơn lẻ trong C++

Vui lòng tìm kiếm trên google

## Tham gia kênh của chúng tôi

Tham gia kênh telegram của chúng tôi để nhận thông tin cập nhật tức thì về khấu hao và các tính năng mới trên HTML, CSS, JavaScript, jQuery, Node. js, PHP và Python

Cho số nguyên n, ta cần tìm dãy số nguyên dương sao cho tất cả các số trong dãy đó là hợp số và độ dài của dãy đó là n. Bạn có thể in bất kỳ phạm vi nào trong trường hợp có nhiều câu trả lời. Hợp số là số nguyên dương có ít nhất một ước khác 1 và chính nó (Nguồn. viwiki)

ví dụ.

```Input : 3
Output : [122, 124]
Explanation 122, 123, 124 are all composite numbers```

Đề nghị thực hành

Phạm vi của các số Composite

Thử nó

Giải pháp là một chút khó khăn. Vì có nhiều câu trả lời có thể, chúng tôi thảo luận về một giải pháp tổng quát ở đây.

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```

Ví dụ cho thuật toán trên

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```

## C++

`// C++ program to find a range of`

`// composite numbers of given length`

`#include `

`using` `namespace` `std;`

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
0

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
1

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
2
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
3
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
2
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
5

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
6

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
8
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
9

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
0_______6_______1
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
2

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
1
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
5

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
6

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
7

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
8

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
2
`[122, 124]`
0_______5_______2
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
5

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
6

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
2
`[122, 124]`
5

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
2
`[122, 124]`
7

`[122, 124]`
8
`[122, 124]`
9 `// C++ program to find a range of`0`// C++ program to find a range of`1 `// C++ program to find a range of`2`// C++ program to find a range of`3`// C++ program to find a range of`4

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
1 `// C++ program to find a range of`6

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
6

`// C++ program to find a range of`8

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
2 `// composite numbers of given length`0

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
6

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
2 `// composite numbers of given length`4

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7`// composite numbers of given length`6

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
1 `// C++ program to find a range of`6

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
6

`#include `1

## Java

`#include `2

`#include `3

`#include `4 `#include `5

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
6

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7`#include `8

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7`using`0
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
2 `using`2_______5_______2
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
5

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
6

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
0_______5_______8 `using`9`namespace`0`namespace`1

`namespace`2

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
1 `namespace`4`// C++ program to find a range of`4

`namespace`6

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
0_______6_______1 `namespace`9`namespace`4`std;`1

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
6

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
7

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7`std;`8

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7`using`0
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
01
`[122, 124]`
0
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
2
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
5

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
6

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
07
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
09
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
10_______5_______11
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
10`// C++ program to find a range of`4

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
07
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
16`namespace`4`// C++ program to find a range of`4

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
07
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
20
`[122, 124]`
9
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
22`// C++ program to find a range of`1
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
24`// C++ program to find a range of`3`std;`1

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
6

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7`// C++ program to find a range of`8

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
33 `using`0
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
01
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
36
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
37
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
38

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
6

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
0_______5_______2
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
43
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
44 `// C++ program to find a range of`4

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
0`// composite numbers of given length`6

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
6

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
6

## Python3

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
51

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
52

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
53

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
54
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
55

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
57_______5_______58 `namespace`4

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
61
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
62_______5_______63
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
64_______5_______65
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
10
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
67
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
68 `namespace`4
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
70

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
0_______5_______57
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
73
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
58
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
62

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
1
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
57

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
79

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
80

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
54
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
82

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
57_______5_______58
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
86
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
68
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
10`namespace`1_______5_______68
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
10

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
93_______5_______58
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
57_______5_______68
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
97
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
98 `namespace`4

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
01
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
65
`[122, 124]`
9
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
68
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
05
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
06
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
68`// C++ program to find a range of`1
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
68
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
05
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
11
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
68`// C++ program to find a range of`3`namespace`1

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
15

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
97
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
58
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
44

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
19

## C#

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
20

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
21

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
22

`using`

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
24

`#include `4

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
26

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
0

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
1

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7`using`0
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
2 `using`2_______5_______2
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
5

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
6

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
0_______5_______8
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
9

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
0_______6_______1
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
2

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
0

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
0_______6_______1
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
49

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
6

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
7

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
8

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7`using`0
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
01
`[122, 124]`
0
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
2
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
5

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
6

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
2
`[122, 124]`
5

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
2
`[122, 124]`
7

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
72
`[122, 124]`
9
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
22

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
75`// C++ program to find a range of`1
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
24`// C++ program to find a range of`3`std;`1

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
6

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7`// C++ program to find a range of`8

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
33 `using`0
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
01
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
89

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
6

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
0_______5_______2 `// composite numbers of given length`4

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
0`// composite numbers of given length`6

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
6

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
6

`[122, 124]`
00

## PHP

`[122, 124]`
01

`[122, 124]`
02

`// composite numbers of given length`

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
0

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
1

`[122, 124]`
06
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
3_______27_______08`namespace`1

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
6

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
8
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
65
`[122, 124]`
08
`[122, 124]`
15

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
0_______6_______1
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
2

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
7
```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
1
`[122, 124]`
08
`[122, 124]`
22
`[122, 124]`
08
`[122, 124]`
24

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
6

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
7

```n = 3
Then a = (n+2)! + 2
a = 5! + 2
a + 1 = 5! + 3
a + 2 = 5! + 4
Here a is divisible by 2
Here a + 1 is divisible by 3
Here a + 2 is divisible by 4
Hence a, a+1, a+2 are all composites```
8

`[122, 124]`
06
`[122, 124]`
0
`[122, 124]`
08`namespace`1

```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
6

`[122, 124]`
33
`[122, 124]`
34
`[122, 124]`
08
`[122, 124]`
36

`[122, 124]`
37
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
58
`[122, 124]`
33
```Let the length of range be n and range starts
from a then, a, a+1, a+2, ...., a+n-1 all should
be composite. So the problem boils down to finding
such 'a'.

If we closely observe p! (where p is a positive
integers) then we will find that, p! has factors of
2, 3, 4, ..., p-1,
Hence if we add i to p! such that 1 < i < p,
then p! + i has a factor i, so p! + i must be
composite. So we end up finding p! + 2, p! + 3,
... p! + p-1 are all composite and continuous
integers forming a range [p! + 2, p! + p-1]
The above range consists of p-2 elements.
For a range of n elements we need to consider (n+2)!

If we take a = (n+2)! + 2,
Then, a + 1 = (n+2)! + 3
Then, a + 2 = (n+2)! + 4
...
Then, a + n-1 = (n+2)! + n+1
Hence,
a = (n+2)! + 2 = 2*3*....*(n+2) + 2
a has 2 as its divisor because (n+2)! and 2
both divides 2
a + 1 = 2*3*....*(n+2) + 3
a + 1 has 3 as its divisor because (n+2)!
and 3 both divides 3
...
a + n-1 = 2*3*....*(n+2) + n+1
a + n-1 has n+1 as its divisor because (n+2)!
and n+1 both divides n+1

Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]```
68
`[122, 124]`
08
`[122, 124]`
42

`[122, 124]`
43
`[122, 124]`
9
`[122, 124]`
45
`[122, 124]`
33
`[122, 124]`
45`// C++ program to find a range of`1
`[122, 124]`
45
`[122, 124]`
37
`[122, 124]`
45`// C++ program to find a range of`3`// C++ program to find a range of`4

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