# 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>
// 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

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