Hi, I wasn't sure whether to post this here or in the pre-calc forum, so apologies if this is in the wrong section. I'm working through Vellemans 'How to prove it' and he gives a proof that if 2|n and 3|n then 6|n (note that a|n means a divides n, just in case this is not standard notation). I think I proved it a different way, but as he did it differently, I'm half assuming my proof is incorrect. His proof is as follows: "Suppose 2 | n and 3 | n. Then we can choose integers j and k such that n=2j and n=3k. Therefore 6(j−k)=6j−6k=3(2j)−2(3k)= 3n − 2n = n, so 6 | n." My "proof" is as follows: "Suppose 2 | n and 3 | n. Then there is an integer k such that (2)(3)k=n. Thus, (6)k=n and therefore 6 | n." If my proof is wrong, then could you tell me why? I was wondering if it's because my proof has something to do with the fact that every number can be written as a product of prime numbers and maybe I need to state this or something. Thanks in advance.