C Programming: Count Integers with Odd digit sum

C Programming Mathematics: Exercise-38 with Solution

Accept a positive integer (n) from the user. Write a C program that counts the number of positive integers from 1 to n whose digit sums are odd.

Example:
Input: n = 5
Integers less than or equal to 5 whose digit sums are odd are 1,3 and 5.
Output: 3
Input: n = 10
Integers less than or equal to 5 whose digit sums are odd are 1, 3, 5, 7, 9 and 10 (1+0 =1)
Output: 6

Test Data:
(5) -> 3
(10) -> 6
(11) -> 6

Sample Solution:

C Code:

#include <stdio.h>

int digit_sum(int n) {
  int d_sum = 0;
  while (n > 0) {
    d_sum += n % 10;
    n /= 10;
  }
  return d_sum;
}

int test(int num) {
  int result = 0;
  for (int i = 1; i <= num; i++) {
    if (digit_sum(i) % 2 != 0) {
      result++;
    }
  }
  return result;
}
int main(void) {
  int n = 5;
  printf("\nIntegers with Odd digit sum from 1 and %d = %d", n, test(n));
  n = 10;
  printf("\nIntegers with Odd digit sum from 1 and %d = %d", n, test(n));
}

Sample Output:

Integers with Odd digit sum from 1 and 5 = 3
Integers with Odd digit sum from 1 and 10 = 6

Flowchart:

C Programming Code Editor:

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