# C/C++ Program to check whether it is possible to make a divisible by 3 number using all digits in an array

Given an array of integers, the task is to find whether it is possible to construct an integer using all the digits of these numbers such that it would be divisible by 3. If it is possible then print “Yes” and if not print “No”.

Examples:

Input : arr[] = {40, 50, 90} Output : Yes We can construct a number which is divisible by 3, for example 945000. So the answer is Yes. Input : arr[] = {1, 4} Output : No The only possible numbers are 14 and 41, but both of them are not divisible by 3, so the answer is No.

`// C++ program to find if it is possible` `// to make a number divisible by 3 using` `// all digits of given array` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `bool` `isPossibleToMakeDivisible(` `int` `arr[], ` `int` `n)` `{` ` ` `// Find remainder of sum when divided by 3` ` ` `int` `remainder = 0;` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `remainder = (remainder + arr[i]) % 3;` ` ` ` ` `// Return true if remainder is 0.` ` ` `return` `(remainder == 0);` `}` ` ` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 40, 50, 90 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `if` `(isPossibleToMakeDivisible(arr, n))` ` ` `printf` `(` `"Yes\n"` `);` ` ` `else` ` ` `printf` `(` `"No\n"` `);` ` ` ` ` `return` `0;` `}` |

**Output:**

Yes

Time Complexity: O(n)

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