Check if Y can be made multiple of X by adding any value to both

Given two numbers X and Y (X ≤ Y), you can select any positive integer Z and add them to both X and Y. The task is to find whether it is possible to make X a multiple of Y.

Examples:

Input: X = 7, Y = 15
Output: Yes

Explanation: We can choose Z = 1 and add them to 7 and 15. Thus, 7 + 1 = 8 is a factor of 15 + 1 = 16.

Input: X = 9, Y = 10

Output: No

Approach: The problem can be solved based on the following idea:

• First, if X = Y, then the answer is obviously “Yes”.
• Otherwise, note that no matter which Z we choose, the difference between X and Y remains constant.
  Let d = Y – X. A valid Z exists if and only if X ≤ d.

Follow the steps mentioned below to implement the above idea:

• First, we find the difference between X and Y let d = Y – X.
• If d = 0 or X ≤ d, then print “Yes”
• Else print “No”

Below is the implementation of the above approach.

C++

#include <bits/stdc++.h>
using namespace std;

// Function to whether it is possible
// to make X a multiple of Y
void check(int X, int Y)
{
    int d = Y - X;
    if (d == 0 || X <= d)
        cout << ("Yes");
    else
        cout << ("No");
}

// Driver code
int main()
{
    int X = 7;
    int Y = 15;

    // Function Call
    check(X, Y);
}

// This code is contributed by garg28harsh.

Java

// Java code to implement the approach

import java.io.*;
import java.util.*;

class GFG {

    // Function to whether it is possible
    // to make X a multiple of Y
    public static void check(int X, int Y)
    {
        int d = Y - X;
        if (d == 0 || X <= d)
            System.out.println("Yes");
        else
            System.out.println("No");
    }

    // Driver code
    public static void main(String[] args)
    {
        int X = 7;
        int Y = 15;

        // Function Call
        check(X, Y);
    }
}

Python3

# Python code to implement the approach

# Function to whether it is possible
# to make X a multiple of Y
def check(X, Y):
    d = Y - X
    if(d == 0 or X <= d):
        print("Yes")
    else:
        print("No")

# Driver code
X = 7
Y = 15

# Function Call
check(X, Y)

# This code is contributed by Pushpesh Raj.

C#

// C# code to implement the approach
using System;
public class GFG {

  // Function to whether it is possible
  // to make X a multiple of Y
  public static void check(int X, int Y)
  {
    int d = Y - X;
    if (d == 0 || X <= d)
      Console.WriteLine("Yes");
    else
      Console.WriteLine("No");
  }

  // Driver code
  public static void Main(string[] args)
  {
    int X = 7;
    int Y = 15;

    // Function Call
    check(X, Y);
  }
}

// This code is contributed by AnkThon

Javascript

// JavaScript code for the above approach

// Function to whether it is possible
// to make X a multiple of Y
function check(X, Y)
{
    let d = Y - X;
    if (d == 0 || X <= d)
        console.log("Yes");
    else
        console.log("No");
}

// Driver code

let X = 7;
let Y = 15;

// Function Call
check(X, Y);

// This code is contributed by Potta Lokesh

Output

Yes

Time Complexity: O(1) 
Auxiliary Space: O(1)

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