Check if X and Y can be made zero by using given operation any number of times

Given two integers X and Y, the task is to check if these two integers can be made equal to 0 by using the given operation any number of times. An operation is described as follows – 

• Choose an arbitrary integer Z.
• Update the values with either of the following values: 
  • X = X – 2*Z and Y = Y – 3*Z
  • X = X – 3*Z and Y = Y – 2*Z

Examples:

Input: X = 6, Y = 9 
Output: Yes 
Explanation:
Operation 1: Choose Z = 3, X = 6 – 2*3 = 0 and Y = 9 – 3*3 = 0 
Since X and Y can be made equal to 0 using 1 operation, the required answer is Yes.
Input: X = 33, Y = 27 
Output: Yes 
Explanation:
Operation 1: Choose Z = 9, X := 33 – 3*9 = 6 and Y := 27 – 2*9 = 9 
Operation 2: Choose Z = 3, X := 6 – 2*3 = 0 and Y := 9 – 3*3 = 0 
Since X and Y can be made equal to 0 using 2 operation, the required answer is Yes. 

Approach: Let’s assume X ? Y. Then the answer is Yes if two following conditions holds: 

• (X + Y) mod 5 = 0: because after each operation the value (X + Y) mod 5 does not change. 
  Let’s assume some arbitrary number Z has been chosen. 
  Therefore, the value (X + Y) will be changed to 
  

((X - 3Z) + (Y - 2Z))

• This is equal to 
  

(X + Y - 5Z)

• For this value to be equal to 0, X + Y = 5Z. Therefore, on taking mod on both the sides, (X + Y) mod 5 has to be equal to 0.
• 3*X >= 2*Y so that the subtraction doesnt make the values of X and Y negative.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach

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

// Function to check if X and Y
// can be made equal to zero by
// using given operation any number of times
void ifPossible(int X, int Y)
{
    if (X > Y)
        swap(X, Y);

    // Check for the two conditions
    if ((X + Y) % 5 == 0 and 3 * X >= 2 * Y)
        cout << "Yes";
    else
        cout << "No";
}

// Driver code
int main()
{
    int X = 33, Y = 27;

    ifPossible(X, Y);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to check if X and Y
// can be made equal to zero by
// using given operation any number of times
static void ifPossible(int X, int Y)
{
    if (X > Y)
        swap(X, Y);

    // Check for the two conditions
    if ((X + Y) % 5 == 0 && 3 * X >= 2 * Y)
        System.out.print("Yes");
    else
        System.out.print("No");
}
static void swap(int x, int y)
{
    int temp = x;
    x = y;
    y = temp;
}

// Driver code
public static void main(String[] args)
{
    int X = 33, Y = 27;

    ifPossible(X, Y);
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach

# Function to check if X and Y
# can be made equal to zero by
# using given operation any number of times
def ifPossible(X, Y):
    if (X > Y):
        X, Y = Y, X

    # Check for the two conditions
    if ((X + Y) % 5 == 0 and 3 * X >= 2 * Y):
        print("Yes")
    else:
        print("No")

# Driver code
X = 33
Y = 27

ifPossible(X, Y)

# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to check if X and Y
// can be made equal to zero by
// using given operation any number of times
static void ifPossible(int X, int Y)
{
    if (X > Y)
        swap(X, Y);

    // Check for the two conditions
    if ((X + Y) % 5 == 0 && 3 * X >= 2 * Y)
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
static void swap(int x, int y)
{
    int temp = x;
    x = y;
    y = temp;
}

// Driver code
public static void Main()
{
    int X = 33, Y = 27;

    ifPossible(X, Y);
}
}

// This code is contributed by Yash_R

Javascript

<script>

// Javascript implementation of the approach

// Function to check if X and Y
// can be made equal to zero by
// using given operation any number of times
function ifPossible(X, Y)
{
    if (X > Y)
    {
        var temp = X;
        X = Y;
        Y =temp;
    }

    // Check for the two conditions
    if ((X + Y) % 5 == 0 && 3 * X >= 2 * Y)
        document.write( "Yes");
    else
        document.write( "No");
}

// Driver code
var X = 33, Y = 27;
ifPossible(X, Y);

// This code is contributed by rutvik_56.
</script>

Output:

Yes

Time Complexity: O(1)

Auxiliary Space: O(1)