Find The Number N, Where (N+X) Divisible By Y And (N-Y) Divisible By X

In this tutorial, we will learn how to determine the number N, where (N + X) divisible by Y and (N-Y) divisible by X.

This is Two numbers, X and Y, are given. Finding the number N(N ≤ 10^18) that meets these two requirements: (N + X) can be divided by Y, and (N - Y) can be divided by X.

Examples:

Input: X = 15, Y = 25
Output: 500

Input: X = 18, Y = 42
Output:780

The following concept can be used to solve the problem.

The number system is a straightforward mathematical idea that we can use.

N + X = Y    ............(i)
N - Y = X     ............(ii)

Normalizing the equation allows us to obtain N could be the same as (X*Y - X + Y). Additionally, it meets these two requirements. The solution is therefore (X * Y - X + Y).

The application of the aforementioned strategy is seen below:

C++ Program 1:

#include <iostream>
using namespace std;
#define int long long int

int FindN(int x, int y)
{
	int N = (x * y) - x + y;

	return N;
}

signed main()
{

	int X = 18, Y = 42;

	cout << FindN(X, Y);
	return 0;
}

Output

C++ Program 2:

#include <iostream>
using namespace std;
#define int long long int

int FindN(int x, int y)
{
	int N = (x * y) - x + y;

	return N;
}

signed main()
{

	int X = 15, Y = 25;

	cout << FindN(X, Y);
	return 0;
}

Output

Java Code 1:

import java.util.*
public class Main
{
public static int FindN(int x, int y)
{
	int N = (x * y) - x + y;

	return N;
}
public static void main(String[] args) {
	int X = 18, Y = 42;

	System.out.println(FindN(X, Y));
}
}

Output

Java Code 2:

import java.util.*
public class Main
{
public static int FindN(int x, int y)
{
	int N = (x * y) - x + y;

	return N;
}
public static void main(String[] args) {
	int X = 15, Y = 25;

	System.out.println(FindN(X, Y));
}
}

Output

C# Code 1:

using System;

public class JTP {
static int FindN(int x, int y)
{
	int N = (x * y) - x + y;
	return N;
}

static public void Main()
{
	int X = 18, Y = 42;
	Console.WriteLine(FindN(X, Y));
}
}

Output

C# Code 2:

 using System;

public class JTP {
static int FindN(int x, int y)
{
	int N = (x * y) - x + y;
	return N;
}

static public void Main()
{
	int X = 18, Y = 42;
	Console.WriteLine(FindN(X, Y));
}
}

Output