Largest even digit number not greater than N in C++

In this tutorial, we are going to write a program that finds the largest number whose digits are all even and not greater than the given n.

Let's see the steps to solve the problem.

Initialise the number n.
Write a loop from i = n .
- Check whether the digits of current number are all even or not.
- If the above condition satisfies, then print the number.
- Else decrement the i.

Example

Let's see the code.

Live Demo

#include <bits/stdc++.h>
using namespace std;
int allDigitsEven(int n) {
   while (n) {
      if ((n % 10) % 2){
         return 0;
      }
      n /= 10;
   }
   return 1;
}
int findLargestEvenNumber(int n) {
   int i = n;
   while (true) {
      if (allDigitsEven(i)) {
         return i;
      }
      i--;
   }
}
int main() {
   int N = 43;
   cout << findLargestEvenNumber(N) << endl;
   return 0;
}

Output

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

Conclusion

If you have any queries in the tutorial, mention them in the comment section.

Hafeezul Kareem

Published on 09-Apr-2021 13:18:44

Previous Page Print Page Next Page