Show that if k divides nm, then k divides gcd(k,n)*gcd(k,m)
This result is easy to think about, but I'm having a hard time making it formal. I'm pretty sure the best way is to look at the prime factorizations of k, n, and m. Each prime which divides k must also divide either n or m. So the product gcd(k,n)*gcd(k,m) will contain all prime factors of k. That seems like hand waving though, I don't know exactly how to make it a solid argument. Any hints would be appreciated. Thanks.