Fundamental Theorem of Arithmetic and Shared Prime Factors

Question

If some integer n is divisible by both 9 and by 6, then it is divisible by 18.

This statement is incredibly straightforward, but I'm struggling to prove it. I can explain why this relationship is true fairly easily (the shared multiples of two integers p and q coincide with the multiples of some integer n, the least common multiple of p and q, because the prime factors of n are accounted for by the prime factors of p and q), but I can't seem to determine how this idea can be included in a more formal proof.

Based on my intuition, it seems directly tied to the fundamental theorem of arithmetic, but I could also be way off. I'd appreciate an indication of where I should go with this proof or how I should get it started, but I would prefer I am not given the complete answer. Thanks.

Answer 1

Using the unique prime factorization is straightforward here.

We know $9=3^2$ and $6=2\cdot 3$.

Now assume a number $n$ is is divisible by both $6$ and $9$. Factorize it using prime powers $n=2^\alpha_23^\alpha_35^\alpha_5\dots$, where the $\alpha_p$ exponents can be zero as well (indeed they are always zero for all but finitely many primes).

So, what can we say about $\alpha_2$ and $\alpha_3$?