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 Show Vì vậy, nếu đầu vào giống như n = 512, thì đầu ra sẽ là 4608 và 4096 bướcĐể giải quyết vấn đề này, chúng tôi sẽ làm theo các bước sau - print 10*n and 9*n. Thí dụChúng 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);
}Đầu vào512 đầu ra5120, 4608 #includeusing tên không gian std; đầu raSử dụng vòng lặp whileTrong 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; đầu raSử dụng vòng lặp do whileTrong 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; đầu raHợp Số Giữa Khoảng Đã ChoTrong 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 raCác số hỗn hợp từ 1 đến 10. 4 6 8 9 10 Kiểm tra xem số đã cho là số nguyên tố hay hợp sốTrong 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 raNhập một số (int). 18 18 là hợp số Lời nhắc nhởXin 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 tôiTham 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++
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 composites0_______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 composites2
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 composites1 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 composites5 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 composites6
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 composites7 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 composites8 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 of0// C++ program to find a range of1 // C++ program to find a range of2// C++ program to find a range of3// C++ program to find a range of4n = 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 composites1 // C++ program to find a range of6n = 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 composites6
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 length0Let 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 length4Let 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 length6Let 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 composites1 // C++ program to find a range of6n = 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 composites6
Java
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 8Let 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 using0 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 using2_______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 composites0_______5_______8 using9namespace0namespace1
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 composites1 namespace4// C++ program to find a range of4
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 composites0_______6_______1 namespace9namespace4std;1Let 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 composites6 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 composites7 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;8Let 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 using0 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]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]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 of4Let 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]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]16 namespace4// C++ program to find a range of4Let 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 of1 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 of3std;1Let 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 composites6 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 of8Let 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 using0 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 composites0_______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 of4n = 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 composites0 // composite numbers of given length6Let 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 composites6 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 composites6 Python3Let 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 namespace4Let 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 namespace4Let 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 composites0_______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 composites1 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 namespace1_______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 namespace4Let 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 composites01 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 composites05 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 composites06 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 of1Let 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 composites05 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 composites11 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 of3namespace1
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 composites15 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 composites19 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 composites20 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 composites21 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 composites22
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 composites24
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 composites26 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 using0 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 using2_______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 composites0_______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 composites0_______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 composites2 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 composites0 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 composites0_______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 composites49 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 composites6 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 composites7 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 composites8 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 using0 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 composites72 [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 composites75 // C++ program to find a range of1 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 of3std;1Let 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 composites6 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 of8Let 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 using0 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 composites89 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 composites0_______5_______2 // composite numbers of given length4n = 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 composites0 // composite numbers of given length6Let 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 composites6 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 composites6
[122, 124]00 PHP[122, 124]01 [122, 124]02
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 namespace1Let 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 composites0_______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 composites2
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 composites1 [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 composites6
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 composites7 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 composites8 [122, 124]06 [122, 124]0 [122, 124]08 namespace1Let 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 of1 [122, 124]45 [122, 124]37 [122, 124]45 // C++ program to find a range of3// C++ program to find a range of4
Cách tìm hợp số trong C?Nếu số đó chia hết cho i thì giá trị của c tăng thêm 1. Khi đó, nếu giá trị của c là 2 có nghĩa là số đó chỉ chia hết cho 2 số (i. e. 1 và chính số đó) nên số đã nhập là số nguyên tố. Nếu không, nó là một hợp số
Các số hỗn hợp từ 21 đến 50 là gì?Giải Các Bài Toán Về Hợp Số Và. Có 34 số tự nhiên từ 1 đến 50 như sau. 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, . 2. Các số hợp số từ 1 đến 100 là gì?
Các số ghép giữa 51 và 70 là gì?Do đó, hợp số giữa đến. 4 , 6 , 8 , 9 , 10 , 12 , 14 , 15 , 16 , 18 , 20 , 21 , 22 , 24 , 25 , 26 , 27 , 28 , 30 , 32 , 33 . Hỏi. . Q.
logic cho hợp số là gì?Hợp số là số nguyên dương không phải số nguyên tố . Nói cách khác, nó có một ước số dương khác với một hoặc chính nó. Một số hợp số đầu tiên là 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, ……… Mọi số nguyên lớn hơn 1 đều là số nguyên tố hoặc hợp số. |
