# GCD and LCM mix question

Suppose $$A,B$$,and $$C$$ are integers greater than or equal to $$2$$. If $$\gcd(A,B)=12, \text{lcm}(A,B)=396$$ and $$\gcd(B,C)= 33$$, what is the $$\gcd(11A,B)$$?

Since $$\gcd(a,b)= 12$$ we have $$a=12x$$ and $$b=12y$$ where $$x,y$$ are relatively prime.
Since $$\gcd(b,c)=33$$ we have $$33\mid b\implies 33\mid 12y\implies 11\mid y\implies y = 11z$$ so, since $$x,z$$ are relativley prime, we have $$\gcd (11a,b) = \gcd (11\cdot 12x, 12\cdot 11z) = 132$$
Notice that we do not need lcm$$(a,b) = 396$$.
$$(11A,B) = (11A,11B,B)=(11\overbrace{(A,B)}^{\large 12},B)=11\cdot 12,\,$$ by $$\,11,12\mid B\,\Rightarrow\,11\cdot 12\mid B$$