Given an array of integers, the task is to find whether it’s 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.

The idea is based on the fact that a number is divisible by 3 iff sum of its digits is divisible by 3. So we simply find sum of array elements. If the sum is divisible by 3, our answer is Yes, else No.

## CPP

`// 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; ` `} ` |

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## Java

`// Java program to find if it is possible ` `// to make a number divisible by 3 using ` `// all digits of given array ` ` ` `import` `java.io.*; ` `import` `java.util.*; ` ` ` `class` `GFG ` `{ ` ` ` `public` `static` `boolean` `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` `); ` ` ` `} ` ` ` ` ` `public` `static` `void` `main (String[] args) ` ` ` `{ ` ` ` `int` `arr[] = { ` `40` `, ` `50` `, ` `90` `}; ` ` ` `int` `n = ` `3` `; ` ` ` `if` `(isPossibleToMakeDivisible(arr, n)) ` ` ` `System.out.print(` `"Yes\n"` `); ` ` ` `else` ` ` `System.out.print(` `"No\n"` `); ` ` ` `} ` `} ` ` ` `// Code Contributed by Mohit Gupta_OMG <(0_o)> ` |

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## Python3

`# Python program to find if it is possible ` `# to make a number divisible by 3 using ` `# all digits of given array ` ` ` `def` `isPossibleToMakeDivisible(arr, n): ` ` ` `# Find remainder of sum when divided by 3 ` ` ` `remainder ` `=` `0` ` ` `for` `i ` `in` `range` `(` `0` `, n): ` ` ` `remainder ` `=` `(remainder ` `+` `arr[i]) ` `%` `3` ` ` ` ` `# Return true if remainder is 0. ` ` ` `return` `(remainder ` `=` `=` `0` `) ` ` ` `# main() ` ` ` `arr ` `=` `[` `40` `, ` `50` `, ` `90` `]; ` `n ` `=` `3` `if` `(isPossibleToMakeDivisible(arr, n)): ` ` ` `print` `(` `"Yes"` `) ` `else` `: ` ` ` `print` `(` `"No"` `) ` ` ` `# Code Contributed by Mohit Gupta_OMG <(0_o)> ` |

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## C#

`// C# program to find if it is possible ` `// to make a number divisible by 3 using ` `// all digits of given array ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `public` `static` `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); ` ` ` `} ` ` ` ` ` `public` `static` `void` `Main () ` ` ` `{ ` ` ` `int` `[]arr = { 40, 50, 90 }; ` ` ` `int` `n = 3; ` ` ` ` ` `if` `(isPossibleToMakeDivisible(arr, n)) ` ` ` `Console.WriteLine(` `"Yes"` `); ` ` ` `else` ` ` `Console.WriteLine(` `"No"` `); ` ` ` `} ` `} ` ` ` `// This code is contributed by vt_m ` |

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## PHP

`<?php ` `// PHP program to find if it is possible ` `// to make a number divisible by 3 using ` `// all digits of given array ` ` ` `function` `isPossibleToMakeDivisible(` `$arr` `, ` `$n` `) ` `{ ` ` ` ` ` `// Find remainder of sum ` ` ` `// when divided by 3 ` ` ` `$remainder` `= 0; ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++) ` ` ` `$remainder` `= (` `$remainder` `+ ` `$arr` `[` `$i` `]) % 3; ` ` ` ` ` `// Return true if remainder is 0. ` ` ` `return` `(` `$remainder` `== 0); ` `} ` ` ` `// Driver code ` `$arr` `= ` `array` `( 40, 50, 90 ); ` `$n` `= sizeof(` `$arr` `); ` `if` `(isPossibleToMakeDivisible(` `$arr` `, ` `$n` `)) ` ` ` `echo` `(` `"Yes\n"` `); ` `else` ` ` `echo` `(` `"No\n"` `); ` ` ` `// This code is contributed by Ajit. ` `?> ` |

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**Output:**

Yes

Time Complexity : O(n)

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