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Given a number x , find y such that x*y + 1 is not a prime

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  • Last Updated : 08 Mar, 2022
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Given a number x, Find y (y > 0) such that x*y + 1 is not a prime.
Examples: 
 

Input : 2
Output : 4

Input : 5
Output : 3

 

Observation: 
 

x*(x-2) + 1 = (x-1)^2 which is not a prime. 
 

Approach : 

For x > 2, y will be x-2 otherwise y will be x+2

 

C++




#include <bits/stdc++.h>
using namespace std;
 
int findY(int x)
{
    if (x > 2)
        return x - 2;
 
    return x + 2;
}
 
// Driver code
int main()
{
    int x = 5;
    cout << findY(x);
    return 0;
}

C




#include <stdio.h>
 
int findY(int x)
{
    if (x > 2)
        return x - 2;
  
    return x + 2;
}
  
// Driver code
int main()
{
    int x = 5;
    printf("%d",findY(x));
    return 0;
}

Java




// JAVA implementation of above approach
 
import java.util.*;
 
class GFG
{
    public static int findY(int x)
    {
        if (x > 2)
            return x - 2;
     
        return x + 2;
    }
 
    // Driver code
    public static void  main(String [] args)
    {
        int x = 5;
        System.out.println(findY(x));
 
    }
     
}
 
 
// This code is contributed
// by ihritik

Python3




# Python3 implementation of above
# approach
 
def findY(x):
 
    if (x > 2):
        return x - 2
     
    return x + 2
 
# Driver code
if __name__=='__main__':
    x = 5
    print(findY(x))
 
# This code is contributed
# by ihritik

C#




// C# implementation of above approach
using System;
 
class GFG
{
public static int findY(int x)
{
    if (x > 2)
        return x - 2;
 
    return x + 2;
}
 
// Driver code
public static void Main()
{
    int x = 5;
    Console.WriteLine(findY(x));
}
}
 
// This code is contributed
// by Subhadeep

PHP




<?php
// PHP implementation of above approach
function findY($x)
{
    if ($x > 2)
        return $x - 2;
 
    return $x + 2;
}
 
// Driver code
$x = 5;
echo (findY($x));
 
// This code is contributed
// by Shivi_Aggarwal
?>

Javascript




<script>
 
// JavaScript implementation of above approach
 
// Function to check whether it is possible
// or not to move from (0, 0) to (x, y)
// in exactly n steps
function findY(x)
{
    if (x > 2)
        return x - 2;
 
    return x + 2;
}
 
// Driver code
 
var x = 5;
document.write(findY(x));
 
// This code is contributed by Ankita saini
 
</script>

Output:  

3

 


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