Help with Proof: GCD(a,bk) = GCD(a,k)

[Hells_Kitchen]

Hi there can someone help me on this one:
If gcd(a,b)=1 then gcd(a,bk) = gcd(a,k)

I have arrived at the conclusion that gcd(a,bk) gcd(a,k) both divide k but from here I do not get anywhere.

I would like some hints on this please.

 

[redmeteos]

Hi, I think it is sufficient to prove that gcd(a,bk) and gcd(a,b) divide each other.

 

[ramsey2879]

Hi there can someone help me on this one:
If gcd(a,b)=1 then gcd(a,bk) = gcd(a,k)

I have arrived at the conclusion that gcd(a,bk) gcd(a,k) both divide k but from here I do not get anywhere.

I would like some hints on this please.

could be that they both equal 1, for instance if k = b but think of expressing both a and k in the form of a product of primes. Since the gcd(a,k) is the product of the prime factors that are common to a and k or 1 if there are no common factors other than 1, and gcd(a,b), which is the product of the prime factors that are common to b and a, is 1, does k*b add any other factors of a?