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# Smallest positive integer X satisfying the given equation

Given two integers N and K, the task is to find the smallest positive integer X satisfying the equation:

(X / K) * (X % K) = N

Examples:

Input: N = 6, K = 3
Output: 11
Explanation:
For X = 11, (11 / 3) * (11 % 3) = 3 * 2 = 6
Therefore, the following equation satisfies.

Input: N = 4, K = 6
Output: 10
Explanation:
For X = 10, (10 / 6) * (10 % 6) = 1 * 4 = 4
Therefore, the following equation satisfies.

Brute Force Approach:

The brute force approach to solve this problem would be to iterate over all positive integers X and check if the given equation is satisfied. If the equation is satisfied for any value of X, we return that value as the smallest positive integer satisfying the equation.

Below is the implementation of the above approach:

## C++

 #include using namespace std; // Function to find out the smallest// positive integer for the equationint findMinSoln(int n, int k){    int x = 1;    while (true) {        if ((x / k) * (x % k) == n) {            return x;        }        x++;    }} // Driver Codeint main(){    int n = 4, k = 6;    cout << findMinSoln(n, k);    return 0;}

## Java

 public class Main {    // Function to find out the smallest    // positive integer for the equation    public static int findMinSoln(int n, int k) {        int x = 1;        while (true) {            if ((x / k) * (x % k) == n) {                return x;            }            x++;        }    }     // Driver Code    public static void main(String[] args) {        int n = 4, k = 6;        System.out.println(findMinSoln(n, k));    }}

## Python3

 # Function to find out the smallest# positive integer for the equationdef findMinSoln(n, k):    x = 1    while True:        if (x // k) * (x % k) == n:            return x        x += 1 # Driver Codeif __name__ == '__main__':    n, k = 4, 6    print(findMinSoln(n, k))

Output

10

Time Complexity: O(X), which can be very large if X is a large number.
Auxiliary Space : O(1)

Approach:

The idea is to observe that, since (X / K) * (X % K) = N, therefore, N will be divisible by p = X % K, which is less than K. Therefore, for all i in the range [1, K) try all values of p where: Below is the implementation of the above approach:

## C++

 // C++ Program to implement// the above approach#include using namespace std; // Function to find out the smallest// positive integer for the equationint findMinSoln(int n, int k){    // Stores the minimum    int minSoln = INT_MAX;     // Iterate till K    for (int i = 1; i < k; i++) {         // Check if n is divisible by i        if (n % i == 0)            minSoln                = min(minSoln, (n / i) * k + i);    }     // Return the answer    return minSoln;} // Driver Codeint main(){    int n = 4, k = 6;    cout << findMinSoln(n, k);}

## Java

 // Java Program to implement// the above approachimport java.util.*;class GFG{  // Function to find out the smallest// positive integer for the equationstatic int findMinSoln(int n, int k){    // Stores the minimum    int minSoln = Integer.MAX_VALUE;      // Iterate till K    for (int i = 1; i < k; i++)    {          // Check if n is divisible by i        if (n % i == 0)            minSoln = Math.min(minSoln, (n / i) * k + i);    }      // Return the answer    return minSoln;}  // Driver Codepublic static void main(String[] args){    int n = 4, k = 6;    System.out.println(findMinSoln(n, k));}} // This code is contributed by Ritik Bansal

## Python3

 # Python3 program to implement# the above approachimport sys # Function to find out the smallest# positive integer for the equationdef findMinSoln(n, k):         # Stores the minimum    minSoln = sys.maxsize;     # Iterate till K    for i in range(1, k):         # Check if n is divisible by i        if (n % i == 0):            minSoln = min(minSoln, (n // i) * k + i);         # Return the answer    return minSoln; # Driver Codeif __name__ == '__main__':         n = 4;    k = 6;         print(findMinSoln(n, k)); # This code is contributed by amal kumar choubey

## C#

 // C# program to implement// the above approachusing System; class GFG{ // Function to find out the smallest// positive integer for the equationstatic int findMinSoln(int n, int k){         // Stores the minimum    int minSoln = int.MaxValue;     // Iterate till K    for (int i = 1; i < k; i++)    {         // Check if n is divisible by i        if (n % i == 0)            minSoln = Math.Min(minSoln,                              (n / i) * k + i);    }     // Return the answer    return minSoln;} // Driver Codepublic static void Main(String[] args){    int n = 4, k = 6;         Console.WriteLine(findMinSoln(n, k));}} // This code is contributed by amal kumar choubey

## Javascript

 

Output:

10

Time Complexity: O(N)
Auxiliary Space: O(1)

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