OldChemE Posted March 8, 2019 Share Posted March 8, 2019 (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, 2019 by OldChemE Link to comment Share on other sites More sharing options...

taeto Posted March 8, 2019 Share Posted March 8, 2019 You are not missing anything. Where does the question come from though? Link to comment Share on other sites More sharing options...

Roamer Posted March 8, 2019 Share Posted March 8, 2019 x can still be 1 Link to comment Share on other sites More sharing options...

mathematic Posted March 8, 2019 Share Posted March 8, 2019 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. Link to comment Share on other sites More sharing options...

studiot Posted March 8, 2019 Share Posted March 8, 2019 (edited) Rethink Edited March 8, 2019 by studiot Rethink Link to comment Share on other sites More sharing options...

OldChemE Posted March 9, 2019 Author Share Posted March 9, 2019 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. Link to comment Share on other sites More sharing options...

studiot Posted March 9, 2019 Share Posted March 9, 2019 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. Link to comment Share on other sites More sharing options...

taeto Posted March 9, 2019 Share Posted March 9, 2019 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? Link to comment Share on other sites More sharing options...

Sensei Posted March 9, 2019 Share Posted March 9, 2019 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} .. ? Link to comment Share on other sites More sharing options...

taeto Posted March 9, 2019 Share Posted March 9, 2019 8 minutes ago, Sensei said: ..maybe instead of xy^{3} there should be (xy)^{3} .. ? That would make it worse... Link to comment Share on other sites More sharing options...

studiot Posted March 10, 2019 Share Posted March 10, 2019 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. Link to comment Share on other sites More sharing options...

taeto Posted March 10, 2019 Share Posted March 10, 2019 (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, 2019 by taeto Link to comment Share on other sites More sharing options...

studiot Posted March 10, 2019 Share Posted March 10, 2019 1 hour ago, taeto said: Theorem 1 (studiot's theorem) Lemma, please. Theorem is a bit grand. 1 Link to comment Share on other sites More sharing options...

taeto Posted March 10, 2019 Share Posted March 10, 2019 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 Link to comment Share on other sites More sharing options...

studiot Posted March 13, 2019 Share Posted March 13, 2019 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 Link to comment Share on other sites More sharing options...

taeto Posted March 14, 2019 Share Posted March 14, 2019 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". Link to comment Share on other sites More sharing options...

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