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".

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

// 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; 
} 

Java

// 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 

Python3

# python program to find if all 
# prime factors of y divide x. 
  
def gcd(a, b): 
    if(b == 0): 
        return a  
    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 

C#

// 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. 

PHP

<?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 
?> 

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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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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