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C++ Program for GCD of more than two (or array) numbers
• Difficulty Level : Expert
• Last Updated : 04 Dec, 2018

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)
```
 `// C++ program to find GCD of two or``// more numbers``#include ``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;``    ``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);``    ``cout << findGCD(arr, n) << endl;``    ``return` `0;``}`
Output:
```2
```

Please refer complete article on GCD of more than two (or array) numbers for more details!

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