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## Main Question or Discussion Point

I forgot how to prove the following, can anyone jog my memory:

Let m,n be natural numbers s.t. gcd(m,n) = 1 and n|k and m|k for some natural number k.

Then nm|k.

I know I'll kick myself when I find out. Thanks.

Let m,n be natural numbers s.t. gcd(m,n) = 1 and n|k and m|k for some natural number k.

Then nm|k.

I know I'll kick myself when I find out. Thanks.