# Divisibility | Practice problem Hackerearth solution

Problem:-

You are provided an array

$A$

of size

$N$

that contains non-negative integers. Your task is to determine whether the number that is formed by selecting the last digit of all the N numbers is divisible by

$10$

.

Note: View the sample explanation section for more clarification.

Input format

• First line: A single integer
$N$

denoting the size of array

$A$

• Second line:
$N$

space-separated integers.

Output format

If the number is divisible by

$10$

, then print

$Yes$

. Otherwise, print

$No$

.

Constraints

$1\le N\le {10}^{5}\phantom{\rule{0ex}{0ex}}0\le A\left[i\right]\le {10}^{5}$

Sample Input
`585 25 65 21 84`
Sample Output
`No`
Time Limit: 1
Memory Limit: 256
Source Limit:
Explanation

Last digit of

$85$

is

$5$

.
Last digit of

$25$

is

$5$

.
Last digit of

$65$

is

$5$

.
Last digit of

$21$

is

$1$

.
Last digit of

$84$

is

$4$

.
Therefore the number formed is

$55514$

which is not divisible by

$10$

.

Code:-

Here I am going to give you two solution first one is on the basis of C language and second one is on the basis of c++ language which you can submit in c++14 and c++17 also

Solution 1 ( C language):-

include <stdio.h>
int main(){
int N = 0;
scanf(“%d”, &N);

long data[N];
for(auto i=0; i<N; i++)
scanf(“%ld”, &data[i]);
char ans[3];

if (data[N1] % 10 == 0)
printf(“Yes”);
else
printf(“No”);
return 0;
}

Solution 2 ( C++ language):-

This solution is based on the c++ language and you can submit ib c++14 and c++17 also.
In this solution first three lines of the main function is only for the decreasing the time of execution of the program..
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);

This is your choice that you want to use this or not but in some cases the code may take more time in execution and that time we need it .

#include<bits/stdc++.h>
using namespace std;
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n,i;
cin>>n;
int a[n] , b[n];
for(i=0;i<n;i++)
{
cin>>a[i];
}
for(i=0;i<n;i++)
{
b[i] = a[i]%10;
}
if(b[n1] == 0)
cout<<“Yesn”;
else
cout<<“Non”;
}

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