# OK-- I'm baffled

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Posted (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 x2Y2  and xy3 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 y2 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 x2 y2 becomes 9x2.    In order for this to have a GCF of 27, x2 must contain at least one factor of 3.

If x2 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 x2 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 by OldChemE

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You are not missing anything. Where does the question come from though?

x can still be 1

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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.

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Posted (edited)

Rethink

Edited by studiot
Rethink

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I appreciate the responses-- but I don't think the question has been answered.  If x = 1 and y = 3 then the GCF of x2y2 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.

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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.

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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?

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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 xy3  there should be (xy)3 .. ?

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8 minutes ago, Sensei said:

..maybe instead of xy3  there should be (xy)3 .. ?

That would make it worse...

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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.

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Posted (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 by taeto

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1 hour ago, taeto said:

Theorem 1 (studiot's theorem)

Lemma, please. Theorem is a bit grand.

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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

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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.﻿

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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".