Sum of digits using recursion

Idea. The last digit is n % 10 and the remaining digits are n / 10. So

sumDigits(n) = (n % 10) + sumDigits(n / 10), and sumDigits(0) = 0 (base case).

Trace for 12345: 5 + sumDigits(1234) = 5 + 4 + sumDigits(123) = 5 + 4 + 3 + 2 + 1 = 15.

#include <stdio.h>

int sumDigits(int n)
{
    if (n == 0)                      /* base case */
        return 0;
    return (n % 10) + sumDigits(n / 10);
}

int main(void)
{
    int n;
    printf("Enter a 5-digit number: ");
    scanf("%d", &n);
    if (n < 10000 || n > 99999)
        printf("Not a 5-digit number\n");
    else
        printf("Sum of digits = %d\n", sumDigits(n));
    return 0;
}

Sample run

Input: 12345

Enter a 5-digit number: Sum of digits = 15