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.

42

Conclusion

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