Is it possible to find GCD, the greatest common divisor of more than?
Yes, of course. For example, you can rewrite GCD of three numbers in terms of GCD on pairs of numbers. \text{gcd}(a, b, c) = \text{gcd}(\text{gcd}(a,b), c) = \text{gcd}(a,\text{gcd}(b,c)) And, of course, you can extend that as far as you like. \text{gcd} and \text{lcm} act like \text{min} and \text{max}. Why? Because they are \text{min} and \text{max} , in a manner of speaking. Specifically, they apply \text{min} and \text{max} to the exponents of the prime factorizations of their arguments. Let's say I have a function f(a, b), which returns the number of times b divides a repeatedly without a remainder. Or, put differently, it returns the largest integer exponent n such that ab^{-n} is an integer. We'll only worry about defining this function for a, b \in \mathbb{Z}, a \geq , b > 1. If you pass in a prime p_i for the second argumenti.e. f(a, p_i)it gives you the exponent of p_i in a's prime factorization. Note: I use the common convention that p_i is the i^{\text{th}} prime. Let's use the number 84 as an example. Its prime factorization is 2^2 \cdot 3 \cdot 7. \begin{array}{rcl} f(84,2) &=& 2\\ f(84,3) &=& 1 \\ f(84,5) &=& \\ f(84,7) &=& 1 \end{array} So how does this relate to \text{gcd} and \text{lcm}? \Large \displaystyle \text{gcd}(a_0, a_1, \dots a_n) = \prod_i p_i^{\text{min}(f(a_0, p_i), f(a_1, p_i), \dots f(a_n, p_i))} \Large \displaystyle \text{lcm}(a_0, a_1, \dots a_n) = \prod_i p_i^{\text{max}(f(a_0, p_i), f(a_1, p_i), \dots f(a_n, p_i))} In English: The exponent on each prime factor of the GCD is the minimum of all the exponents for that prime in the prime factorizations of the arguments to GCD. The exponent on each prime factor of the LCM is the maximum of all the exponents for that prime in the prime factorizations of the arguments to LCM. That's a mouthful. It sounds complicated, but it isn't, really. Work a few GCD and LCM problems in terms of prime factorizations, though, and hopefully it should click.