N-th Number Whose Sum of Digits is Ten in C++

The numbers whose digits sum is equal to 10 are

19, 28, 37, 46, 55, 64, 73, 82, 91, etc..,

If you observe the series, each number is incremented by 9. There are numbers in the above sequence whose digits sum does not equal 10 while incrementing by 9. But, you will get all the numbers whose digits sum is equal to 10.

So, we can have a loop that increments by 9 and checks for digits sum and finds the n-th number. Let's see some examples

Inputs

3
7

Outputs

37
73

Algorithm

• Initialise the number n
• Initialise a counter to 0.
• Write a loop that iterates from 19
  • If the current number digits sum is 10, increment the counter by 1.
  • If the counter is equal to n, then return the current number.
  • Increment the iterative variable by 9.

Implementation

Following is the implementation of the above algorithm in C++

#include <bits/stdc++.h>
using namespace std;
int findNthNumber(int n) {
   int count = 0, i = 19;
   while (true) {
      int sum = 0;
      for (int number = i; number > 0; number = number / 10) {
         sum = sum + number % 10;
      }
      if (sum == 10) {
         count++;
      }
      if (count == n) {
         return i;
      }
      i += 9;
   }
   return -1;
}
int main() {
   int n = 7;
   cout << findNthNumber(7) << endl;
   return 0;
}

Output

If you run the above code, then you will get the following result.

73