I know how to solve for 2 nos.
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I'm quoting the source cited below.
The GCD of three or more numbers equals the product of the prime factors common to all the numbers,[13] which can be calculated by taking the GCDs of pairs of numbers.[14] For example,
GCD(a, b, c) = GCD(a, GCD(b, c)) = GCD(GCD(a, b), c) = GCD(GCD(a, c), b).
Thus, Euclid's algorithm, which computes the GCD of two integers, suffices to calculate the GCD of arbitrarily many integers.
The GCD of three or more numbers equals the product of the prime factors common to all the numbers,[13] which can be calculated by taking the GCDs of pairs of numbers.[14] For example,
GCD(a, b, c) = GCD(a, GCD(b, c)) = GCD(GCD(a, b), c) = GCD(GCD(a, c), b).
Thus, Euclid's algorithm, which computes the GCD of two integers, suffices to calculate the GCD of arbitrarily many integers.