# Check if the number formed by the last digits of N numbers is divisible by 10 or not

Given an array arr[] of size N consisting of non-zero positive integers. The task is to determine whether the number that is formed by selecting the last digits of all the numbers is divisible by 10 or not. If the number is divisible by 10, then print Yes otherwise print No.
Examples:

Input: arr[] = {12, 65, 46, 37, 99}
Output: No
25679 is not divisible by 10.
Input: arr[] = {24, 37, 46, 50}
Output: Yes
4760 is divisible by 10.

Approach: In order for an integer to be divisible by 10, it must be ending with a 0. So, the last element of the array will decide whether the number formed will be divisible by 10 or not. If the last digit of the last element is 0 then print Yes otherwise print No.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function that returns true if the``// number formed by the last digits of``// all the elements is divisible by 10``bool` `isDivisible(``int` `arr[], ``int` `n)``{``    ``// Last digit of the last element``    ``int` `lastDigit = arr[n - 1] % 10;` `    ``// Number formed will be divisible by 10``    ``if` `(lastDigit == 0)``        ``return` `true``;` `    ``return` `false``;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 12, 65, 46, 37, 99 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``if` `(isDivisible(arr, n))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;``    ` `class` `GFG``{` `// Function that returns true if the``// number formed by the last digits of``// all the elements is divisible by 10``static` `boolean` `isDivisible(``int` `arr[], ``int` `n)``{``    ``// Last digit of the last element``    ``int` `lastDigit = arr[n - ``1``] % ``10``;` `    ``// Number formed will be divisible by 10``    ``if` `(lastDigit == ``0``)``        ``return` `true``;` `    ``return` `false``;``}` `// Driver code``static` `public` `void` `main ( String []arg)``{``    ``int` `arr[] = { ``12``, ``65``, ``46``, ``37``, ``99` `};``    ``int` `n = arr.length;` `    ``if` `(isDivisible(arr, n))``        ``System.out.println(``"Yes"``);``    ``else``        ``System.out.println(``"No"``);``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the approach ` `# Function that returns true if the ``# number formed by the last digits of ``# all the elements is divisible by 10 ``def` `isDivisible(arr, n) : ` `    ``# Last digit of the last element ``    ``lastDigit ``=` `arr[n ``-` `1``] ``%` `10``; ` `    ``# Number formed will be divisible by 10 ``    ``if` `(lastDigit ``=``=` `0``) : ``        ``return` `True``; ` `    ``return` `False``; ` `# Driver code ``if` `__name__ ``=``=` `"__main__"` `: ` `    ``arr ``=` `[ ``12``, ``65``, ``46``, ``37``, ``99` `]; ``    ``n ``=` `len``(arr); ` `    ``if` `(isDivisible(arr, n)) :``        ``print``(``"Yes"``); ``    ``else` `:``        ``print``(``"No"``); ` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{` `// Function that returns true if the``// number formed by the last digits of``// all the elements is divisible by 10``static` `bool` `isDivisible(``int` `[]arr, ``int` `n)``{``    ``// Last digit of the last element``    ``int` `lastDigit = arr[n - 1] % 10;` `    ``// Number formed will be divisible by 10``    ``if` `(lastDigit == 0)``        ``return` `true``;` `    ``return` `false``;``}` `// Driver code``static` `public` `void` `Main(String []arg)``{``    ``int` `[]arr = { 12, 65, 46, 37, 99 };``    ``int` `n = arr.Length;` `    ``if` `(isDivisible(arr, n))``        ``Console.WriteLine(``"Yes"``);``    ``else``        ``Console.WriteLine(``"No"``);``}``}``    ` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:
`No`

Time Complexity: O(1)

Auxiliary Space: O(1)

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