If GCD of a list of numbers is 1, is it a necessary condition that GCD of at least one pair of numbers from the list should be 1?

Question

Suppose our numbers are {2, 6, 3}. GCD (2, 6, 3) = 1, GCD (2, 6) = 2, GCD (6, 3) = 3, but GCD(2, 3) = 1

If GCD(a,b,c) = p, GCD(a,b) = q, GCD(b,c) = r, GCD(c,a) = s, is it possible that p = 1 and (q != 1 && r != 1 && s != 1) ?

Is there any intuition behind this scenario ?

Answer 1

Consider the numbers $6,10,15$. Their $\gcd$ is $1$, but the $\gcd$ of any pair is greater than $1$.

More generally, we can produce a list $x_1,x_2,\dots,x_n$ of integers with $\gcd$ $1$ such that the $\gcd$ of any $n-1$ of them is greater than $1$. Let $p_1,p_2,\dots,p_n$ be distinct primes, and let $N$ be their product. Let $x_k=\frac{N}{p_k}$.

Answer 2

Look at the prime factorization of the numbers. The gcd is 1 if there is no prime number p which is simultaneously a factor of all numbers. But if there is more than one prime number involved, then each prime number can be a factor of two of the numbers. Take three primes p, q, r, and the numbers pq, pr, qr. Each two numbers have a common prime factor, but no prime is factor of all three.