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Given two numbers X and Y. the duty is to seek out the quantity N (N ≤ 10^18) which satisfies these two circumstances:

  • (N + X) is divisible by Y
  • (N – Y) is divisible by X


Enter: X = 18, Y = 42

Enter: X = 100, Y = 200
Output: 500

Strategy: The issue may be solved based mostly on the next thought:

We will implement the straightforward math idea of the quantity system.

  • N + X = Y    …………(i)
  • N – Y = X     …………(ii)

Normalizes the equation we will get N could also be equal to (X*Y – X + Y). Additionally, it’s satisfying these two circumstances. So, the reply is (X * Y – X + Y).

Under is the implementation of the above method:


// C++ code for the above method:
#embody <iostream>
utilizing namespace std;
#outline int lengthy lengthy int

// Perform to seek out N
int FindN(int x, int y)

    // Mathematical equation from method
    int N = (x * y) - x + y;

    // Return N
    return N;

// Driver code
signed primary()

    int X = 18, Y = 42;

    // Perform name
    cout << FindN(X, Y);
    return 0;

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

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