# GCD of more than two (or array) numbers

Given an array of numbers, find GCD of the array elements. In a previous post we find GCD of two number.

Examples:

```Input  : arr[] = {1, 2, 3}
Output : 1

Input  : arr[] = {2, 4, 6, 8}
Output : 2
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

The GCD of three or more numbers equals the product of the prime factors common to all the numbers, but it can also be calculated by repeatedly taking the GCDs of pairs of numbers.

```gcd(a, b, c) = gcd(a, gcd(b, c))
= gcd(gcd(a, b), c)
= gcd(gcd(a, c), b)
```

For an array of elements, we do following.

```result = arr[0]
For i = 1 to n-1
result = GCD(result, arr[i])
```

Below is C++ implementation of above idea.

## C++

```// C++ program to find GCD of two or
// more numbers
#include<bits/stdc++.h>
using namespace std;

// Function to return gcd of a and b
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}

// Function to find gcd of array of
// numbers
int findGCD(int arr[], int n)
{
int result = arr[0];
for (int i=1; i<n; i++)
result = gcd(arr[i], result);

return result;
}

// Driven code
int main()
{
int arr[] = {2, 4, 6, 8, 16};
int n = sizeof(arr)/sizeof(arr[0]);
cout << findGCD(arr, n) << endl;
return 0;
}
```

## Java

```// Java program to find GCD of two or
// more numbers

public class GCD
{
// Function to return gcd of a and b
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}

// Function to find gcd of array of
// numbers
static int findGCD(int arr[], int n)
{
int result = arr[0];
for (int i=1; i<n; i++)
result = gcd(arr[i], result);

return result;
}

public static void main(String[] args)
{
int arr[] = {2, 4, 6, 8, 16};
int n = arr.length;
System.out.println(findGCD(arr, n));
}
}

// This code is contributed by Saket Kumar
```

## Python

```
# GCD of more than two (or array) numbers
# This function implements the Euclidian algorithm to find H.C.F. of two number

def find_gcd(x, y):
while(y):
x, y = y, x % y

return x

l = [2, 4, 6, 8, 16]

num1=l[0]
num2=l[1]
gcd=find_gcd(num1,num2)

for i in range(2,len(l)):
gcd=find_gcd(gcd,l[i])

print(gcd)

# Code contributed by Mohit Gupta_OMG
```

Output:

```2
```

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