My Math Forum Match Question sorrry if posted in the wrong section

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 August 27th, 2009, 01:57 PM #1 Newbie   Joined: Aug 2009 Posts: 1 Thanks: 0 Match Question sorrry if posted in the wrong section can anyone tell me how to figure this out (without just plugging in the values experimentally)? If x and y are positive integers such that the greatest common factor of x^2y^2 and xy^3 is 45, then which of the following could y equal? a. 45 b. 15 c. 9 d. 5 e. 3
 August 29th, 2009, 10:35 AM #2 Senior Member   Joined: Feb 2009 Posts: 172 Thanks: 5 Re: Match Question sorrry if posted in the wrong section I'm not a number guy, but think that's how you do it. You must have $x\neq y$ cause if they were equal you would have that $x^2y^2=xy^3=y^4$ and then the greatest common factor would be $y^4$. but $\sqrt[4]{45}$ is not an integer. Now, the greatest common factor of $x^2y^2$ and $xy^3$ is $xy^2$. Then you have that $xy^2=45=5\cdot 3^2$ wich gives you that $y=3$. If you find any mistake tell me
 September 1st, 2009, 03:56 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,379 Thanks: 2011 Since gcd(x²y², xy³) = xy²(gcd(x, y)) = 45 = 3²(5), 45 is a multiple of y², and so y = 1 or 3. Hence, of the alternative values for y listed, only 3 is possible.
 September 1st, 2009, 06:49 PM #4 Senior Member   Joined: Aug 2008 From: Blacksburg VA USA Posts: 344 Thanks: 6 Math Focus: primes of course Re: Match Question sorrry if posted in the wrong section I plugged into my latest fav site wolframalpha.com and it confirms the earlier post input box: gcd(x²y², xy³)=45 it reports integer solutions at x,y = (45,1), (5,3) and shows a nice plot via the power of Mathematica behind it

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# if x and y are positive integers such that the greatest common factor of x^2y^2 and xy^3 is 45

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