# GCD of factorials of two numbers

• Difficulty Level : Hard
• Last Updated : 06 Apr, 2021

Given two numbers m and n. Find the GCD of the their factorial.
Examples :

Input : n = 3, m = 4
Output : 6
Explanation:
Factorial of n = 1 * 2 * 3 = 6
Factorial of m = 1 * 2 * 3 * 4 = 24
GCD(6, 24) = 6.

Input : n = 9, m = 5
Output : 20
Explanation:
Factorial of n = 1 * 2 * 3 *4 * 5 * 6 * 7 * 8
* 9 = 362880
Factorial of m = 1 * 2 * 3 * 4 * 5 = 120
GCD(362880, 120) = 120

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A simple solution is to find factorials of both numbers. Then find GCD of the computed factorials.
An efficient solution is based on the fact that GCD of two factorials is equal to smaller factorial (note that factorials have all terms common).
Below is the implementation of above approach.

## C++

 // CPP program to find GCD of factorial of two// numbers.#include using namespace std; int factorial(int x){    if (x <= 1)        return 1;    int res = 2;    for (int i = 3; i <= x; i++)        res = res * i;    return res;} int gcdOfFactorial(int m, int n){    return factorial(min(m, n));} int main(){    int m = 5, n = 9;    cout << gcdOfFactorial(m, n);    return 0;}

## Java

 // Java program to find GCD of factorial// of two numbers.public class FactorialGCD{     static int factorial(int x){    if (x <= 1)        return 1;    int res = 2;    for (int i = 3; i <= x; i++)        res = res * i;    return res;} static int gcdOfFactorial(int m, int n){    int min = m < n ? m : n;    return factorial(min);}     /* Driver program to test above functions */    public static void main (String[] args)    {        int m = 5, n = 9;                 System.out.println(gcdOfFactorial(m, n));    }} // This code is contributed by Prerna Saini

## Python

 # Python code to find GCD of factorials of# two numbers.import math def gcdOfFactorial(m, n) :    return math.factorial(min(m, n)) # Driver codem = 5n = 9print(gcdOfFactorial(m, n))

## C#

 // C# program to find GCD of factorial// of two numbers.using System; public class GFG{          static int factorial(int x)    {        if (x <= 1)            return 1;                     int res = 2;                 for (int i = 3; i <= x; i++)            res = res * i;                     return res;    }          static int gcdOfFactorial(int m, int n)    {        int min = m < n ? m : n;        return factorial(min);    }      /* Driver program to test above functions */    public static void Main()    {                 int m = 5, n = 9;                  Console.WriteLine(gcdOfFactorial(m, n));    }}  // This code is contributed by Anant Agarwal.



## Javascript



Output :

120

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