# Minimum possible value of max(A, B) such that LCM(A, B) = C

Given an integer C, the task is to find the minimum possible value of max(A, B) such that LCM(A, B) = C.

Examples:

Input: C = 6
Output: 3
max(1, 6) = 6
max(2, 3) = 3
and min(6, 3) = 3

Input: C = 9
Output: 9

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: An approach to solve this problem is to find all the factors of the given number using the approach discussed in this article and then find the pair (A, B) that satisfies the given conditions and take the overall minimum of the maximum of these pairs.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the LCM of a and b ` `int` `lcm(``int` `a, ``int` `b) ` `{ ` `    ``return` `(a / __gcd(a, b) * b); ` `} ` ` `  `// Function to return the minimized value ` `int` `getMinValue(``int` `c) ` `{ ` `    ``int` `ans = INT_MAX; ` ` `  `    ``// To find the factors ` `    ``for` `(``int` `i = 1; i <= ``sqrt``(c); i++) { ` ` `  `        ``// To check if i is a factor of c and ` `        ``// the minimum possible number ` `        ``// satisfying the given conditions ` `        ``if` `(c % i == 0 && lcm(i, c / i) == c) { ` ` `  `            ``// Update the answer ` `            ``ans = min(ans, max(i, c / i)); ` `        ``} ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `c = 6; ` ` `  `    ``cout << getMinValue(c); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `Solution ` `{ ` `    ``// Recursive function to return gcd of a and b  ` `    ``static` `int` `__gcd(``int` `a, ``int` `b)  ` `    ``{  ` `        ``// Everything divides 0  ` `        ``if` `(a == ``0``)  ` `        ``return` `b;  ` `        ``if` `(b == ``0``)  ` `        ``return` `a;  ` `         `  `        ``// base case  ` `        ``if` `(a == b)  ` `            ``return` `a;  ` `         `  `        ``// a is greater  ` `        ``if` `(a > b)  ` `            ``return` `__gcd(a - b, b);  ` `        ``return` `__gcd(a, b - a);  ` `    ``} ` `     `  `    ``// Function to return the LCM of a and b ` `    ``static` `int` `lcm(``int` `a, ``int` `b) ` `    ``{ ` `        ``return` `(a / __gcd(a, b) * b); ` `    ``} ` ` `  `    ``// Function to return the minimized value ` `    ``static` `int` `getMinValue(``int` `c) ` `    ``{  ` `        ``int` `ans = Integer.MAX_VALUE; ` ` `  `        ``// To find the factors ` `        ``for` `(``int` `i = ``1``; i <= Math.sqrt(c); i++)  ` `        ``{ ` ` `  `            ``// To check if i is a factor of c and ` `            ``// the minimum possible number ` `            ``// satisfying the given conditions ` `            ``if` `(c % i == ``0` `&& lcm(i, c / i) == c) ` `            ``{ ` ` `  `                ``// Update the answer ` `                ``ans = Math.min(ans, Math.max(i, c / i)); ` `            ``} ` `        ``} ` `        ``return` `ans; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `c = ``6``; ` ` `  `        ``System.out.println(getMinValue(c)); ` `    ``} ` `} ` ` `  `// This code is contributed by Arnab Kundu `

## Python3

 `# Python implementation of the approach ` `import` `sys ` ` `  `# Recursive function to return gcd of a and b ` `def` `__gcd(a, b): ` `     `  `    ``# Everything divides 0 ` `    ``if` `(a ``=``=` `0``): ` `        ``return` `b; ` `    ``if` `(b ``=``=` `0``): ` `        ``return` `a; ` ` `  `    ``# base case ` `    ``if` `(a ``=``=` `b): ` `        ``return` `a; ` ` `  `    ``# a is greater ` `    ``if` `(a > b): ` `        ``return` `__gcd(a ``-` `b, b); ` `    ``return` `__gcd(a, b ``-` `a); ` ` `  `# Function to return the LCM of a and b ` `def` `lcm(a, b): ` `    ``return` `(a ``/` `__gcd(a, b) ``*` `b); ` ` `  `# Function to return the minimized value ` `def` `getMinValue(c): ` `    ``ans ``=` `sys.maxsize; ` ` `  `    ``# To find the factors ` `    ``for` `i ``in` `range``(``1``, ``int``(``pow``(c, ``1``/``2``)) ``+` `1``): ` ` `  `        ``# To check if i is a factor of c and ` `        ``# the minimum possible number ` `        ``# satisfying the given conditions ` `        ``if` `(c ``%` `i ``=``=` `0` `and` `lcm(i, c ``/` `i) ``=``=` `c): ` ` `  `            ``# Update the answer ` `            ``ans ``=` `min``(ans, ``max``(i, c ``/` `i)); ` `    ``return` `int``(ans); ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``c ``=` `6``; ` ` `  `    ``print``(getMinValue(c)); ` `     `  `# This code is contributed by 29AjayKumar `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG ` `{  ` `    ``// Recursive function to return gcd of a and b  ` `    ``static` `int` `__gcd(``int` `a, ``int` `b)  ` `    ``{  ` `        ``// Everything divides 0  ` `        ``if` `(a == 0)  ` `        ``return` `b;  ` `        ``if` `(b == 0)  ` `        ``return` `a;  ` `         `  `        ``// base case  ` `        ``if` `(a == b)  ` `            ``return` `a;  ` `         `  `        ``// a is greater  ` `        ``if` `(a > b)  ` `            ``return` `__gcd(a - b, b);  ` `        ``return` `__gcd(a, b - a);  ` `    ``}  ` `     `  `    ``// Function to return the LCM of a and b  ` `    ``static` `int` `lcm(``int` `a, ``int` `b)  ` `    ``{  ` `        ``return` `(a / __gcd(a, b) * b);  ` `    ``}  ` ` `  `    ``// Function to return the minimized value  ` `    ``static` `int` `getMinValue(``int` `c)  ` `    ``{  ` `        ``int` `ans = ``int``.MaxValue;  ` ` `  `        ``// To find the factors  ` `        ``for` `(``int` `i = 1; i <= Math.Sqrt(c); i++)  ` `        ``{  ` ` `  `            ``// To check if i is a factor of c and  ` `            ``// the minimum possible number  ` `            ``// satisfying the given conditions  ` `            ``if` `(c % i == 0 && lcm(i, c / i) == c)  ` `            ``{  ` ` `  `                ``// Update the answer  ` `                ``ans = Math.Min(ans, Math.Max(i, c / i));  ` `            ``}  ` `        ``}  ` `        ``return` `ans;  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `c = 6;  ` ` `  `        ``Console.WriteLine(getMinValue(c));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```3
```

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