OldChemE 38 Posted March 8 (edited) In my spare time I work as a substitute teacher. In High School math today, students were working on a standardized practice test to prepare for the ACT exam. The following question baffled me. Given x and y are positive integers, such that both x^{2}Y^{2}^{ }and xy^{3} have a greatest common factor (GCF) of 27, which of the following could be a value of y? The available answers are 81, 27, 18, 9 and 3 I ruled out 81, 27, 18 and 9 as values of y because in all cases y^{2} would be greater than 27, so the GCF could not be 27. It appears the intended answer is y = 3. However, if y = 3, then x^{2} y^{2} becomes 9x^{2}. In order for this to have a GCF of 27, x^{2} must contain at least one factor of 3. If x^{2} contains a single (or odd number) factor of 3, then x must contain a factor of square root of 3, and x will not be a positive integer, which violates the problem statement. If x^{2} contains two (or an even number) factors of 3, then x will contain one or more factors of three, and the GCF will be at least 81, which violates the problem statement. What am I missing?? Any thoughts? Thanks Edited March 8 by OldChemE 0 Share this post Link to post Share on other sites

taeto 50 Posted March 8 You are not missing anything. Where does the question come from though? 0 Share this post Link to post Share on other sites

mathematic 77 Posted March 8 The question is "could be a value of y". All the other choices would force the greatest common factor to be greater than 27. For x=3 (times anything which is not divisible by 3) , 27 is fine for the greatest common factor. This gives a restriction on possible x values only. 0 Share this post Link to post Share on other sites

studiot 1609 Posted March 8 (edited) Rethink Edited March 8 by studiot Rethink 0 Share this post Link to post Share on other sites

OldChemE 38 Posted March 9 I appreciate the responses-- but I don't think the question has been answered. If x = 1 and y = 3 then the GCF of x^{2}y^{2} becomes 9, which does not satisfy the constraints of the problem. If x = 3, then the GCF of both terms becomes 81, which does not satisfy the constraints of the problem. If x = square root of 3, then the GCF constraint is satisfied, but x then is not a positive integer, which violates the other constraint of the problem. I'm still trying to understand how both constraints of the problem (x and y both positive integers, and the GCF of both terms = 27) can be satisfied with the available answers. 0 Share this post Link to post Share on other sites

studiot 1609 Posted March 9 Just a note to observe that the problem itself is satisfied by X = 27 Y=1 So maybe they got their list of answers wrong by omitting 1. 0 Share this post Link to post Share on other sites

taeto 50 Posted March 9 1 hour ago, studiot said: So maybe they got their list of answers wrong by omitting 1. It seems like if there had not been any list of possibilities given beforehand, then \(y=1\) would be the unique solution? 0 Share this post Link to post Share on other sites

Sensei 834 Posted March 9 On 3/8/2019 at 12:30 PM, taeto said: You are not missing anything. Where does the question come from though? ..maybe instead of xy^{3} there should be (xy)^{3} .. ? 0 Share this post Link to post Share on other sites

taeto 50 Posted March 9 8 minutes ago, Sensei said: ..maybe instead of xy^{3} there should be (xy)^{3} .. ? That would make it worse... 0 Share this post Link to post Share on other sites

studiot 1609 Posted March 10 On 3/9/2019 at 10:54 AM, taeto said: It seems like if there had not been any list of possibilities given beforehand, then y=1 would be the unique solution? I haven't proved that but If the GCF = g where g is any integer except 1 then Y = 1 x = g is a solution of the problem. 0 Share this post Link to post Share on other sites

taeto 50 Posted March 10 (edited) Okay then. But I like to use \(\gcd\) (greatest common divisor) instead of GCF, since that is the standard, and otherwise I will invariably make typos. Unfortunately "factor" is a little ambiguous. Assume that we have the information for positive integers \(x,y,\) that \(x^2y^2\) and \(xy^3\) have \(\gcd(x^2y^2,xy^3)=g\) as their GCF, for some prescribed positive integer \(g.\) The question is what \(y\) can possibly be? Theorem 1 (studiot's theorem). \(y=1\) is always possible. Proof. Let \(x=g.\) Then for \(y=1\) we get \(\gcd(x^2y^2,xy^3) = \gcd(g^2,g) = g.\) So \(y=1\) is possible. Theorem 2. \(y= 2\) is possible if and only if \(g\) is either \(4\) times an odd number or \(8\) times an even number. Proof. Assume \(y=2.\) Then for each \(x\) we have \(\gcd(x^2y^2,xy^3) = \gcd(4x^2,8x) = 4x\gcd(x,2).\) If \(x\) is odd then \(\gcd(x,2)=1,\) which implies \(g = \gcd(x^2y^2,xy^3)= 4x,\) which is \(4\) times an odd number. If \(x\) is even, then \(\gcd(x,2)=2,\) so \(g = \gcd(x^2y^2,xy^3)= 8x,\) eight times an even number. The converses follow by setting \(x=g/4\) if \(g\) is \(4\) times an odd number, and setting \(x=g/8\) if \(g\) is \(8\) times an even number. End of proof. Who will volunteer to do Theorems 3,4,5,...? Edited March 10 by taeto 0 Share this post Link to post Share on other sites

studiot 1609 Posted March 10 1 hour ago, taeto said: Theorem 1 (studiot's theorem) Lemma, please. Theorem is a bit grand. 1 Share this post Link to post Share on other sites

taeto 50 Posted March 10 2 hours ago, studiot said: Lemma, please. Theorem is a bit grand. It isn't a Lemma if it is not useful for anything further 0 Share this post Link to post Share on other sites

studiot 1609 Posted March 13 On 3/10/2019 at 9:51 PM, taeto said: It isn't a Lemma if it is not useful for anything further I have already proposed studiot's law elsewhere: Quote It also conforms to Studiot's law that fact is stranger than the most fanciful fiction. https://www.scienceforums.net/topic/106314-what-warps-space/?tab=comments#comment-993433 0 Share this post Link to post Share on other sites

taeto 50 Posted March 14 On 3/13/2019 at 11:54 AM, studiot said: I believe that you are ahead of Euler by now. He had hundreds of theorems and lemmas, but no "laws". 0 Share this post Link to post Share on other sites