Check if a number is divisible by all prime divisors of another number

Given two integers. We need to find if the first number x is divisible by all prime divisors of y.

Examples :

Input  : x = 120, y = 75
Output : Yes
Explanation :
120 = (2^3)*3*5
75  = 3*(5^2)
120 is divisible by both 3 and 5 which 
are the prime divisors of 75. Hence, 
answer is "Yes".

Input  :  x = 15, y = 6
Output : No
Explanation : 
15 = 3*5.
 6 = 2*3,
15 is not divisible by 2 which is a 
prime divisor of 6. Hence, answer 
is "No".

A simple solution is to find all prime factors of y. For every prime factor, check if it divides x or not.

An efficient solution is based on below facts.
1) if y == 1, then it no prime divisors. Hence answer is “Yes”
2) We find GCD of x and y.
      a) If GCD == 1, then clearly there are no common divisors of x and y, hence answer is “No”.
      b) If GCD > 1, the GCD contains prime divisors which divide x also. Now, we have all unique prime divisor if and only if y/GCD has such unique prime divisor. So we have to find uniqueness for pair (x, y/GCD) using recursion.

C++

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// CPP program to find if all prime factors
// of y divide x.
#include <bits/stdc++.h>
using namespace std;
  
// Returns true if all prime factors of y 
// divide x.
bool isDivisible(int x, int y)
{
    if (y == 1)
        return true;
  
    if (__gcd(x, y) == 1)
        return false;
    return isDivisible(x, y / gcd);
}
  
// Driver Code
int main()
{
    int x = 18, y = 12;
    if (isDivisible(x, y))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}

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Java

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// Java program to find if all 
// prime factors of y divide x.
class Divisible 
{
    public static int gcd(int a, int b) { 
      return b == 0 ? a : gcd(b, a % b); }
      
    // Returns true if all prime factors
    // of y divide x.
    static boolean isDivisible(int x, int y)
    {
        if (y == 1)
            return true;
              
        int z = gcd(x, y);
      
        if (z == 1)
            return false;
      
        return isDivisible(x, y / z);
    }
  
    // Driver program to test above functions
    public static void main(String[] args) 
    {
        int x = 18, y = 12;
        if (isDivisible(x, y))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
// This code is contributed by Prerna Saini

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Python3

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# python program to find if all
# prime factors of y divide x.
  
def gcd(a, b):
    if(b == 0):
        return
    else:
        return gcd(b, a % b)
      
# Returns true if all prime 
# factors of y divide x.
def isDivisible(x,y):
      
    if (y == 1):
        return 1
  
    z = gcd(x, y);
      
    if (z == 1):
        return false;
      
    return isDivisible(x, y / z);
  
# Driver Code
x = 18
y = 12
if (isDivisible(x, y)):
    print("Yes")
else:
    print("No")
  
# This code is contributed by Sam007

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C#

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// C# program to find if all 
// prime factors of y divide x.
using System;
  
class GFG {
      
    public static int gcd(int a, int b)
    
        return b == 0 ? a : gcd(b, a % b);
    }
      
    // Returns true if all prime factors
    // of y divide x.
    static bool isDivisible(int x, int y)
    {
        if (y == 1)
            return true;
              
        int z = gcd(x, y);
      
        if (z == 1)
            return false;
      
        return isDivisible(x, y / z);
    }
  
    // Driver program to test above functions
    public static void Main() 
    {
        int x = 18, y = 12;
          
        if (isDivisible(x, y))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
  
// This code is contributed by vt_m.

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PHP

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<?php
// PHP program to find if all 
// prime factors of y divide x.
  
function gcd ($a, $b
{
    return $b == 0 ? $a
        gcd($b, $a % $b);
  
// Returns true if all prime 
// factors of y divide x.
function isDivisible($x, $y)
{
    if ($y == 1)
        return true;
      
    $z = gcd($x, $y);
      
    if ($z == 1)
        return false;
      
    return isDivisible($x, $y / $z);
}
  
// Driver Code
$x = 18;
$y = 12;
if (isDivisible($x, $y))
    echo "Yes";
else
    echo "No";
  
// This code is contributed by Sam007
?>

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Output :

Yes

Time Complexity:Time complexity for calculating GCD is O(log min(x, y)), and recursion will terminate after log y steps because we are reducing it by a factor greater than one. Overall Time complexity: O(log2y)

This article is contributed by Harsha Mogali. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : vt_m, Sam007



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