*Write a program to find the sum of cube of digits of a number.*

For example,

Input12345Output225Explaination1^{3}+ 2^{3}+ 3^{3}+ 4^{3}+ 5^{3}= 225

**Approach**

The solution to this problem is very simple. We will simply using **% ( MOD operator )** to read each digit of the input number one by one and store their sum in a temporary variable.

**Steps**

- Let
*N*be the input number. - Initialize a variable
*sum = 0*. - Repeat the following steps while
*N > 0*.- Set
*temp = N % 10*. This stores the rightmost digit of*N*in*temp*. - Set
*N = N / 10*. This step removes the rightmost digit of*N*. - Set
*sum = sum + temp * temp * temp*.

- Set
- The sum of the cube of digits of the input number is stored in
*sum*.

```
#include <stdio.h>
#include <math.h>
// returns sum of cube of digits of n
int sum_of_cube(int n) {
int sum = 0;
int temp;
while( n > 0 ){
temp = n % 10; // store rightmost digit of n in temp
n = n / 10; // remove rightmost digit of n
sum = sum + temp * temp * temp;
}
return sum;
}
int main() {
int n;
printf("Enter a Number: ");
scanf("%d", &n);
printf("Sum of the cube of digits of %d = %d", n, sum_of_cube(n));
}
```

**Output**

```
Enter a Number: 12341
Sum of the cube of digits of 12341 = 101
```

**Read**

NidhiWell written!