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Find minimum x such that (x % k) * (x / k) == n
• Difficulty Level : Medium
• Last Updated : 16 Mar, 2021

Given two positive integers n and k. Find minimum positive integer x such that the (x % k) * (x / k) == n, where % is the modulus operator and / is the integer division operator.
Examples:

```Input : n = 4, k = 6
Output :10
Explanation : (10 % 6) * (10 / 6) = (4) * (1) = 4 which is equal to n

Input : n = 5, k = 5
Output : 26```

Naive Solution : A simple approach is to run a while loop until we find a solution which satisfies the given equation, but this would be very slow.
Efficient Solution : The key idea here is to notice that the value of (x % k) lies in the range [1, k – 1]. (0 is not included, since we can’t divide n by (x % k) when it is zero). Now, we need to find the largest possible number in the range that divides n and hence the given equation becomes x = (n * k) / (x % k) + (x % k).
Note : (x % k) is added to the answer since for the current value of modulus (x % k), it must not be contradicting that on one hand x is such that the remainder upon dividing by k is (x % k) and on the other x is (n * k) / (x % k) whose remainder is simply zero when we divide this value by k.
Below is the implementation of the above approach.

## C++

 `// CPP Program to find the minimum``// positive X such that the given``// equation holds true``#include ``using` `namespace` `std;` `// This function gives the required``// answer``int` `minimumX(``int` `n, ``int` `k)``{``    ``int` `ans = INT_MAX;` `    ``// Iterate over all possible``    ``// remainders``    ``for` `(``int` `rem = k - 1; rem > 0; rem--) {` `        ``// it must divide n``        ``if` `(n % rem == 0)``            ``ans = min(ans, rem + (n / rem) * k);``    ``}``    ``return` `ans;``}` `// Driver Code to test above function``int` `main()``{``    ``int` `n = 4, k = 6;``    ``cout << minimumX(n, k) << endl;` `    ``n = 5, k = 5;``    ``cout << minimumX(n, k) << endl;``    ``return` `0;``}`

## Java

 `// Java Program to find the minimum``// positive X such that the given``// equation holds true``class` `Solution``{``// This function gives the required``// answer``static` `int` `minimumX(``int` `n, ``int` `k)``{``    ``int` `ans =Integer.MAX_VALUE;`` ` `    ``// Iterate over all possible``    ``// remainders``    ``for` `(``int` `rem = k - ``1``; rem > ``0``; rem--) {`` ` `        ``// it must divide n``        ``if` `(n % rem == ``0``)``            ``ans = Math.min(ans, rem + (n / rem) * k);``    ``}``    ``return` `ans;``}`` ` `// Driver Code to test above function``public` `static` `void` `main(String args[])``{``    ``int` `n = ``4``, k = ``6``;``    ``System.out.println( minimumX(n, k));`` ` `    ``n = ``5``; k = ``5``;``    ``System.out.println(minimumX(n, k));``    ` `}``}``//contributed by Arnab Kundu`

## Python3

 `# Python 3 program to find the minimum positive``# x such that the given equation holds true` `# This function gives the required answer``def` `minimumX(n, k):``    ` `    ` `    ``ans ``=` `10` `*``*` `18``    ` `    ``# Iterate over all possible remainders``    ``for` `i ``in` `range``(k ``-` `1``, ``0``, ``-``1``):``        ``if` `n ``%` `i ``=``=` `0``:``            ``ans ``=` `min``(ans, i ``+` `(n ``/` `i) ``*` `k)``    ``return` `ans` `# Driver Code``n, k ``=` `4``, ``6` `print``(minimumX(n, k))` `n, k ``=` `5``, ``5` `print``(minimumX(n, k))` `# This code is contributed``# by Mohit Kumar`

## C#

 `// C# Program to find the minimum``// positive X such that the given``// equation holds true` `using` `System;` `public` `class` `GFG{``    ``// This function gives the required``// answer``static` `int` `minimumX(``int` `n, ``int` `k)``{``    ``int` `ans =``int``.MaxValue;` `    ``// Iterate over all possible``    ``// remainders``    ``for` `(``int` `rem = k - 1; rem > 0; rem--) {` `        ``// it must divide n``        ``if` `(n % rem == 0)``            ``ans = Math.Min(ans, rem + (n / rem) * k);``    ``}``    ``return` `ans;``}` `// Driver Code to test above function``    ``static` `public` `void` `Main (){``        ``int` `n = 4, k = 6;``        ``Console.WriteLine( minimumX(n, k));` `        ``n = 5; k = 5;``        ``Console.WriteLine(minimumX(n, k));``    ` `}``}``//This code is contributed by Sachin.`

## PHP

 ` 0; ``\$rem``--)``    ``{` `        ``// it must divide n``        ``if` `(``\$n` `% ``\$rem` `== 0)``            ``\$ans` `= min(``\$ans``, ``\$rem` `+``                      ``(``\$n` `/ ``\$rem``) * ``\$k``);``    ``}``    ``return` `\$ans``;``}` `// Driver Code``\$n` `= 4 ;``\$k` `= 6 ;` `echo` `minimumX(``\$n``, ``\$k``), ``"\n"` `;` `\$n` `= 5 ;``\$k` `= 5 ;` `echo` `minimumX(``\$n``, ``\$k``) ;` `// This code is contributed by Ryuga``?>`

## Javascript

 ``
Output:
```10
26```

Time Complexity : O(k), where k is the given positive integer.

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