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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 $\gcd(11A,B)$?

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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$(a,b) = 396$.

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$(11A,B) = (11A,11B,B)=(11\overbrace{(A,B)}^{\large 12},B)=11\cdot 12,\,$ by $\,11,12\mid B\,\Rightarrow\,11\cdot 12\mid B$

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