Minimum value that divides one number and divisible by other

Given two integer p and q, the task is to find the minimum possible number x such that q % x = 0 and x % p = 0. If the conditions aren’t true for any number then print -1.

Examples:

Input: p = 3, q = 99
Output: 3
99 % 3 = 0
3 % 3 = 0

Input: p = 2, q = 7
Output: -1



Approach: If a number x satisfies the given condition then it’s obvious that q will be divided by p i.e. q % p = 0 because x is a multiple of p and q is a multiple of x.
So the minimum possible value of x will be the GCD of p and q and when q is not divisible by p then no number will satisfy the given condition.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimum valid number
// that satisfies the given conditions
int minValidNumber(int p, int q)
{
    // If possible
    if (q % p == 0)
        return __gcd(p, q);
    else
        return -1;
}
  
// Driver Code
int main()
{
    int p = 2, q = 6;
    cout << minValidNumber(p, q);
    return 0;
}

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Java

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//Java  implementation of the approach
  
import java.io.*;
  
class GFG {
      
    // Function to calculate gcd
    static int __gcd(int a, int b)
    {
          
        // Everything divides 0 
        if (a == 0 || b == 0)
            return 0;
      
        // base case
        if (a == b)
            return a;
      
        // a is greater
        if (a > b)
            return __gcd(a - b, b);
              
        return __gcd(a, b - a);
    }
  
// Function to return the minimum valid number
// that satisfies the given conditions
static int minValidNumber(int p, int q)
{
    // If possible
    if (q % p == 0)
        return __gcd(p, q);
    else
        return -1;
}
  
// Driver Code
    public static void main (String[] args) {
        int p = 2, q = 6;
        System.out.print(minValidNumber(p, q));
  
  
    // THis code is contributed by  Sachin.
    }
}

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Python3

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# Python3 implementation of the approach 
from math import gcd
  
# Function to return the minimum 
# valid number that satisfies the
# given conditions 
def minValidNumber(p, q) :
  
    # If possible 
    if (q % p == 0) : 
        return gcd(p, q) 
    else :
        return -1
  
# Driver Code 
if __name__ == "__main__" :
  
    p, q = 2, 6
    print(minValidNumber(p, q))
  
# This code is contributed by Ryuga

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
    // Function to calculate gcd
    static int __gcd(int a, int b)
    {
           
        // Everything divides 0 
        if (a == 0 || b == 0)
            return 0;
       
        // base case
        if (a == b)
            return a;
       
        // a is greater
        if (a > b)
            return __gcd(a - b, b);
               
        return __gcd(a, b - a);
    }
  
// Function to return the minimum valid number
// that satisfies the given conditions
static int minValidNumber(int p, int q)
{
    // If possible
    if (q % p == 0)
        return __gcd(p, q);
    else
        return -1;
}
  
// Driver Code
public static void Main()
{
    int p = 2, q = 6;
    Console.Write(minValidNumber(p, q));
}
}
  
// THis code is contributed
// by Mukul Singh

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PHP

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<?php
// Php implementation of the approach
  
function gcd($a, $b
  
    // Everything divides 0 
    if($b == 0) 
  
        return $a
  
    return gcd($b , $a % $b); 
  
// Function to return the minimum valid 
// number that satisfies the given conditions
function minValidNumber($p, $q)
{
    // If possible
    if ($q % $p == 0)
        return gcd($p, $q);
    else
        return -1;
}
  
// Driver Code
$p = 2; 
$q = 6;
echo minValidNumber($p, $q);
  
// This code is contributed by ita_c
?>

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Output:

2


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Improved By : Ryuga, Ita_c, Code_Mech, Sach_Code