3-digit Armstrong Number
Write a program to check if a given 3-digit number X is an Armstrong number or not.
Note: A number is an Armstrong number if the number is equal to the sum of the Nth power of its digits.
Input
The first line is an integer X.
Output
The output should be a single line containing True if it satisfies the given conditions, False in all other cases.
Explanation
In the given example
X = 371, The number of digits in 371 is 3. So, 33 + 73 + 13 = 371. Hence,
371 is an Armstrong number.
So, the output should be
True.
Sample Input 1
371
Sample Output 1
True
Sample Input 2
351
Sample Output 2
False
3 digit Armstrong number
n = input['Enter number here: ']
s = int[n[0]]**3+int[n[1]]**3+int[n[2]]**3
if s == int[n]:
print[n, 'is a 3 digit armstrong number']
else:
print[n, 'isn\'t a 3 digit armstrong number']
Enter number here: 153
153 is a 3 digit armstrong number
Enter number here: 155
155 isn't a 3 digit armstrong numberLearn more about our help with Assignments: Python
3-digit Armstrong Number
Write a Python program of 3-digit Armstrong Number. It consists of two test cases
The below link contains 3-digit Armstrong Number - Question, explanation and test cases
//drive.google.com/file/d/1TDj5IcRLuj8MFif4X8tqX9KBoZrr4nO-/view?usp=sharing
We need all test caese can be come while code was running
A positive integer is called an Armstrong number of order n if
abcd... = an + bn + cn + dn + ...
For example,
153 = 1*1*1 + 5*5*5 + 3*3*3 // 153 is an Armstrong number.
Visit this page to learn how you can check whether a number is an Armstrong number or not in Python.
Source Code
# Program to check Armstrong numbers in a certain interval
lower = 100
upper = 2000
for num in range[lower, upper + 1]:
# order of number
order = len[str[num]]
# initialize sum
sum = 0
temp = num
while temp > 0:
digit = temp % 10
sum += digit ** order
temp //= 10
if num == sum:
print[num]
Output
153 370
Here, we have set the lower limit 100 in variable lower and upper limit 2000 in variable upper. We have used for loop to iterate from variable lower to upper. In iteration, the value of lower is increased by 1 and checked whether it is an Armstrong number or not.
You can change the range and test out by changing the variables lower and upper. Note, the variable lower should be lower than upper for this program to work properly.
An integer is referred to as the Armstrong numbers of the order n. When it's each digit is separated, cubed, and added together then it will result in a sum that is the same as the number, [i.e., pqrs... = pn + qn + rn + sn +...].
If there is an Armstrong number with three digits, the total of cubes in each number is equal to the number in itself.
Example:
In this tutorial, we will learn how we can find Armstrong numbers between two given integers by using Python language.
For example:
The method we use below is easy. We look through all numbers within the range. For each number, we first determine the number of digits within it. Make the total number of decimal digits of the current number be the number "n". Then we calculate the sum of the" n-th" power of each digit. If the sum is greater than "K", then we record the result.
Code:
Output:
Please enter the lower range of the integer: 1000 Please enter the upper range of the integer: 10000 The Armstrong Numbers Between Two Given Integers: 1634 8208 9474
Conclusion
In this tutorial, we have discussed how to print Armstrong numbers between two given integers using Python programming.
Given a number x, determine whether the given number is Armstrong number or not.
A positive integer of n digits is called an Armstrong number of order n [order is number of digits] if.
abcd… = pow[a,n] + pow[b,n] + pow[c,n] + pow[d,n] + ….
Example:
Input : 153
Output : Yes
153 is an Armstrong number.
1*1*1 + 5*5*5 + 3*3*3 = 153
Input : 120
Output : No
120 is not a Armstrong number.
1*1*1 + 2*2*2 + 0*0*0 = 9
Input : 1253
Output : No
1253 is not a Armstrong Number
1*1*1*1 + 2*2*2*2 + 5*5*5*5 + 3*3*3*3 = 723
Input : 1634
Output : Yes
1*1*1*1 + 6*6*6*6 + 3*3*3*3 + 4*4*4*4 = 1634
Approach: The idea is to first count number digits [or find order]. Let the number of digits be n. For every digit r in input number x, compute rn. If sum of all such values is equal to n, then return true, else false.
C++
#include
using namespace std;
int power[int x, unsigned int y]
{
if [y == 0]
return
1;
if [y % 2 == 0]
return power[x, y / 2] * power[x, y / 2];
return x * power[x, y / 2] * power[x, y / 2];
}
int order[int x]
{
int n = 0;
while [x] {
n++;
x = x / 10;
}
return n;
}
bool isArmstrong[int x]
{
int n = order[x];
int temp = x, sum = 0;
while
[temp] {
int r = temp % 10;
sum += power[r, n];
temp = temp / 10;
}
return [sum == x];
}
int main[]
{
int x = 153;
cout
Javascript
function NthArmstrong[n]
{
let count = 0;
for[let i = 1; i 0]
{
rem = num % 10;
sum = sum + Math.pow[rem, digit];
num = parseInt[num / 10, 10];
}
if [i == sum]
count++;
if [count == n]
return i;
}
return n;
}
let n = 12;
document.write[NthArmstrong[n]];
C
#include
#include
#include
int NthArmstrong[int n]
{
int count = 0;
for [int i = 1; i 0] {
rem = num % 10;
sum = sum + pow[rem, digit];
num = num / 10;
}
if
[i == sum]
count++;
if [count == n]
return i;
}
}
int main[]
{
int n = 12;
printf["%d", NthArmstrong[n]];
return 0;
}
Using Numeric Strings:
When considering the number as a string we can access any digit we want and perform operations
C++
#include
using namespace std;
string armstrong[int n]{
string number = to_string[n];
n = number.length[];
long long output = 0;
for[char i : number]
output = output + [long]pow[[i - '0'], n];
if [output == stoll[number]]
return["True"];
else
return["False"];
}
int main[]
{
cout