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

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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 <iostream>
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




<script>
// Javascript implementation of the approach
 
// Function that returns true if the
// number formed by the last digits of
// all the elements is divisible by 10
function isDivisible(arr, n)
{
    // Last digit of the last element
    let lastDigit = arr[n - 1] % 10;
 
    // Number formed will be divisible by 10
    if (lastDigit == 0)
        return true;
 
    return false;
}
 
// Driver code
    let arr = [ 12, 65, 46, 37, 99 ];
    let n = arr.length;
 
    if (isDivisible(arr, n))
        document.write("Yes");
    else
        document.write("No");
 
</script>


Output: 

No

 

Time Complexity: O(1)

Auxiliary Space: O(1)



Last Updated : 13 Mar, 2022
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