OK-- I'm baffled

[OldChemE]

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 March 8, 2019 by OldChemE

[taeto]

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

[Roamer]

x can still be 1

[mathematic]

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.

[studiot]

Rethink

 

 

Edited March 8, 2019 by studiot
Rethink

[OldChemE]

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.

[studiot]

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.

[taeto]

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?

[Sensei]

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 .. ?

[taeto]

8 minutes ago, Sensei said:

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

That would make it worse...

[studiot]

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.

[taeto]

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

[studiot]

1 hour ago, taeto said:

Theorem 1 (studiot's theorem)

Lemma, please. Theorem is a bit grand.

[taeto]

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 

[studiot]

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

 

 

[taeto]

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".