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?
Junior 2014 Cox's Bazar Q:9
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
- Raiyan Jamil
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Re: Junior 2014 Cox's Bazar Q:9
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 $1$.
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 $1$.
"Questions we can't answer are far better than answers we can't question"
Re: Junior 2014 Cox's Bazar Q:9
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,$p=(x^{2}y,xy^{2})=xy(x,y)$.
$\therefore$ $xy(x,y)=3xy$.$\therefore$ $(x,y)=3$.
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,$p=(x^{2}y,xy^{2})=xy(x,y)$.
$\therefore$ $xy(x,y)=3xy$.$\therefore$ $(x,y)=3$.
"Questions we can't answer are far better than answers we can't question"