C Program to check if a number is divisible by sum of its digits

Given a number n we have to check whether the sum of its digits divide the number n or not. To find out we have to sum all the numbers starting from the unit place and then divide the number with the final sum.

Like we have a number "521" so we have to find the sum of its digit that will be "5 + 2 + 1 = 8" but 521 is not divisible by 8 without leaving any remainder.

Let's take another example "60" where "6+0 = 6" which can divide 60 and will not leave any remainder.

Syntax

int isDivisible(long int num);

Algorithm

The approach used below is as follows −

• Take input number
• Extract each digit from the unit place and add it to a sum variable which should be initially zero
• Check if the original number is divisible by the sum of digits
• Return the result

Example

#include <stdio.h>

// This function will check
// whether the given number is divisible
// by sum of its digits
int isDivisible(long int num) {
    long int temp = num;
    // Find sum of digits
    int sum = 0;
    while (num) {
        int k = num % 10;
        sum = sum + k;
        num = num / 10;
    }
    // check if sum of digits divides num
    if (temp % sum == 0)
        return 1;
    return 0;
}

int main() {
    long int num1 = 55;
    long int num2 = 12;
    
    printf("Testing %ld: ", num1);
    if(isDivisible(num1))
        printf("Yes<br>");
    else
        printf("No<br>");
    
    printf("Testing %ld: ", num2);
    if(isDivisible(num2))
        printf("Yes<br>");
    else
        printf("No<br>");
        
    return 0;
}

Testing 55: No
Testing 12: Yes

How It Works

• For 55: Sum of digits = 5 + 5 = 10; 55 ÷ 10 = 5.5 (not divisible)
• For 12: Sum of digits = 1 + 2 = 3; 12 ÷ 3 = 4 (divisible)

Conclusion

This program efficiently checks divisibility by extracting each digit using modulo operation and building the sum. The time complexity is O(log n) where n is the input number.