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Check if X and Y can be made zero by using given operation any number of times
  • Last Updated : 23 Mar, 2020

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

Yes

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