Find minimum x such that (x % k) * (x / k) == n

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 that 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.

Steps to solve the problem:

•  Initialize a variable ans to the maximum integer value.
• Iterate over all possible remainders from k-1 to 1 using a for loop with a decrementing step size.
•  If the remainder divides n, calculate the cost of cutting the ribbon into pieces of length k or shorter as rem + (n / rem) * k and 4.    store it in the ans variable.
•  return the minimum value of ans.

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 <bits/stdc++.h>
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

<?php
// PHP Program to find the minimum
// positive X such that the given
// equation holds true

// This function gives the required
// answer
function minimumX($n, $k)
{
    $ans = PHP_INT_MAX;

    // Iterate over all possible
    // remainders
    for ($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
$n = 4 ;
$k = 6 ;

echo minimumX($n, $k), "\n" ;

$n = 5 ;
$k = 5 ;

echo minimumX($n, $k) ;

// This code is contributed by Ryuga
?>

Javascript

<script>

    // Javascript Program to find the minimum
    // positive X such that the given
    // equation holds true

    // This function gives the required
    // answer
    function minimumX(n, k)
    {
        let ans = Number.MAX_VALUE;

        // Iterate over all possible
        // remainders
        for (let 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
    let n = 4, k = 6;
    document.write(minimumX(n, k) + "</br>");

    n = 5, k = 5;
    document.write(minimumX(n, k));

</script>

Output:

10
26

Time Complexity: O(k), where k is the given positive integer.
Auxiliary Space: O(1), since no extra space has been taken.