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Check if the square of a number is divisible by K or not

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  • Last Updated : 12 Nov, 2021
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Given two integers, X and K, the task is to find if X2 is divisible by K or not. Here, both K and X can lie in the range [1,1018].

Examples:  

Input: X = 6, K = 9 
Output: YES 
Explanation: 
Since 62 is equal to 36, which is divisible by 9.

Input: X = 7, K = 21 
Output: NO 
Explanation: 
Since 72 is equal to 49, which is not divisible by 21. 
 

Approach: 
As mentioned above, X can lie in the range [1,1018], so we can not directly check if X2 is divisible by K or not as it can cause a memory overflow. Hence the efficient approach is to first calculate the Greatest Common Divisor of X and K. After the GCD is calculated, we will take it as the maximum portion which we can divide from X and K. Reduce K by GCD and check if X is divisible by the reduced K or not.

Below is the implementation of above approach: 

C++




// C++ implementation to
// check if the square of X is
// divisible by K
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return if
// square of X is divisible
// by K
void checkDivisible(int x, int k)
{
    // Finding gcd of x and k
    int g = __gcd(x, k);
 
    // Dividing k by their gcd
    k /= g;
 
    // Check for divisibility of X
    // by reduced K
    if (x % k == 0) {
        cout << "YES\n";
    }
    else {
        cout << "NO\n";
    }
}
 
// Driver Code
int main()
{
    int x = 6, k = 9;
    checkDivisible(x, k);
 
    return 0;
}

Java




// Java implementation to
// check if the square of X is
// divisible by K
 
class GFG{
 
// Function to return if
// square of X is divisible
// by K
static void checkDivisible(int x, int k)
{
    // Finding gcd of x and k
    int g = __gcd(x, k);
 
    // Dividing k by their gcd
    k /= g;
 
    // Check for divisibility of X
    // by reduced K
    if (x % k == 0)
    {
        System.out.print("YES\n");
    }
    else
    {
        System.out.print("NO\n");
    }
}
static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);    
}
 
// Driver Code
public static void main(String[] args)
{
    int x = 6, k = 9;
    checkDivisible(x, k);
}
}
 
// This code is contributed by gauravrajput1

Python3




# Python3 implementation to
# check if the square of X is
# divisible by K
 
from math import gcd
 
# Function to return if
# square of X is divisible
# by K
def checkDivisible(x, k):
     
    # Finding gcd of x and k
    g = gcd(x, k)
 
    # Dividing k by their gcd
    k //= g
 
    # Check for divisibility of X
    # by reduced K
    if (x % k == 0):
        print("YES")
    else:
        print("NO")
 
# Driver Code
if __name__ == '__main__':
     
    x = 6
    k = 9
    checkDivisible(x, k);
     
# This code is contributed by Bhupendra_Singh

C#




// C# implementation to check
// if the square of X is
// divisible by K
using System;
 
class GFG{
 
// Function to return if
// square of X is divisible
// by K
static void checkDivisible(int x, int k)
{
     
    // Finding gcd of x and k
    int g = __gcd(x, k);
 
    // Dividing k by their gcd
    k /= g;
 
    // Check for divisibility of X
    // by reduced K
    if (x % k == 0)
    {
        Console.Write("YES\n");
    }
    else
    {
        Console.Write("NO\n");
    }
}
 
static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);    
}
 
// Driver Code
public static void Main(String[] args)
{
    int x = 6, k = 9;
     
    checkDivisible(x, k);
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
 
// Javascript implementation to
// check if the square of X is
// divisible by K
 
// Return gcd of two numbers
function gcd(a, b)
{
    return b == 0 ? a : gcd(b, a % b);
}
 
// Function to return if
// square of X is divisible
// by K
function checkDivisible(x, k)
{
     
    // Finding gcd of x and k
    var g = gcd(x, k);
 
    // Dividing k by their gcd
    k /= g;
 
    // Check for divisibility of X
    // by reduced K
    if (x % k == 0)
    {
        document.write("YES");
    }
    else
    {
        document.write("NO");
    }
}
 
// Driver code
var x = 6, k = 9;
checkDivisible(x, k);
 
// This code is contributed by Ankita saini
    
</script>

Output: 

YES

 

Time Complexity: O(log(max(x, k)))

Auxiliary Space;:O(1)


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