C Programming: Count Integers with Odd digit sum

Last update on December 20 2024 13:06:44 (UTC/GMT +8 hours)

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>

// Function to calculate the sum of digits of an integer 'n'
int digit_sum(int n) {
  int d_sum = 0;
  while (n > 0) {
    d_sum += n % 10; // Extracts the last digit of 'n' and adds it to d_sum
    n /= 10; // Removes the last digit by integer division
  }
  return d_sum; // Returns the sum of digits
}

// Function to count the number of integers with odd digit sum from 1 to 'num'
int test(int num) {
  int result = 0;
  for (int i = 1; i <= num; i++) {
    if (digit_sum(i) % 2 != 0) { // Checks if the digit sum of 'i' is odd
      result++; // Increments the counter if the digit sum is odd
    }
  }
  return result; // Returns the count of integers with odd digit sum
}

// Main function
int main(void) {
  int n = 5;
  printf("\nIntegers with Odd digit sum from 1 and %d = %d", n, test(n)); // Displaying the count of integers with odd digit sum up to 'n'

  n = 10;
  printf("\nIntegers with Odd digit sum from 1 and %d = %d", n, test(n)); // Displaying the count of integers with odd digit sum up to 'n'
}

Sample Output:

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

Flowchart:

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