Junior 2014 Cox's Bazar Q:9

[Raiyan Jamil]

x and y are two positive integers. The GCD and LCM of (x^2).y and x.(y^2) are p and q
respectively. If p2=27q then what is the GCD of x and y?

[tanmoy]

Please use latex in writing equation.By the way,here is the solution:
The GCD of $x^{2}y$ and $xy^{2}$ is $xy$.The LCM of $x^{2}y$ and $xy^{2}$ is $x^{2}y^{2}$.I think,there is a problem in the question.It may be $q^{2}=27p$.Because,If $p^{2}=27q$,then $x^{2}y^{2}=27x^{2}y^{2}$ that is impossible.Now,
$x^{4}y^{4}=27xy$
Or, $x^{3}y^{3}=3^{3}$
Or,$xy=3$
$\because $, $x$ and $y$ both positive integer,so,one of them is $3$ and other is $1$.So,GCD of $x$,$y$ is $1$.

[tanmoy]

My previous solution is not correct .There was a big fault.I am really sorry for the mistake . Here is the correct solution:
Given that,$p^{2}=27q$.So,$p^{3}=27pq$
Now,$pq=x^{3}y^{3}$.$\therefore$ $p^{3}=27x^{3}y^{3}$.$p=3xy$.
Now,$p=(x^{2}y,xy^{2})=xy(x,y)$.
$\therefore$ $xy(x,y)=3xy$.$\therefore$ $(x,y)=3$.