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Carla
 
 

OG - Quant Review PS - #169

by Carla Mon May 14, 2007 11:20 am

Question:


If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is:

a)6
b)12
c)24
d)36
e)48

The answer is b.


I broke 72 down into its prime factors and made a prime box for n^2 that includes: {2, 2, 2, 3, 3}

I then wanted to break the prime box of n^2 into two prime boxes for n*n

This is where I was not clear...

I wanted to distribute the elements of n^2 's prime box into two boxes - one box for each n.

So I put {2, 3} in each box for each n.

The issue then is that I have an extra 2. I was not sure where to put it.. so I just added an additional two so that each n prime box contains {2, 2, 3}

However I was not sure if in fact I should have also included the third 2 into each prime box... to create the n box with {2,2,2,3}

Is there a specific method here? How do you know what to include in the boxes for n?

Thanks,
Carla
esledge
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by esledge Tue May 15, 2007 5:45 pm

Quick citation note: This is from the Official Guide for GMAT Quantitative Review. For copyright reasons, we must cite full source name.

Hi Carla,

You are on exactly the right track. Since n is an integer, n^2 must be a perfect square. Thus, the prime factors of n would show up as pairs of those same prime factors in the prime box of n^2.

Think of these prime factors of n^2 as socks fresh out of the dryer. The left socks come from one of the n's and the right socks come from the other n. Even though the n^2 prime box has an unmatched factor of 2, you know the other 2 must be there...like an orphan sock in the laundry--you know the other one is around somewhere. In other words, that unmatched 2 is in n^2's prime box--it must have come from n.

So, your prime box for n should include a 2 and 3 from the complete pairs and the orphan 2, for a prime box of {2,2,3}.
Emily Sledge
Instructor
ManhattanGMAT
Jeff
 
 

prime factos

by Jeff Tue May 15, 2007 6:17 pm

This is one of those problems that you might be able to solve quickly be inspection. You need the smallest perfect square that is divisible by 72. 72 itself does not work, so try the next multiple : 144. If you recognize that 144 is 12^2, then you can mark your answer choice and move on. Otherwise the method that Emily described above will get you there with a little more time and effort.
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by dbernst Wed May 16, 2007 1:29 pm

Good work all! Just one more thing to point out: this can be identified as a higher level question not only by the mathematical content but also by the phrasing of the question itself. By using unconventional wording - "the largest positive integer that must divide n" - rather than the more straightforward "the largest positive integer that must divide evenly into n," the test makers are trying to confuse you with the terminology as well!
vectorSpace
 
 

My thoughts on this question

by vectorSpace Sat Jul 07, 2007 7:35 am

Hello All!

My interpretation of this question was slightly different: and I ended up with a different answer. Here is my reasoning:

"The largest positive integer that must divide n" MUST be n itself. (This is true for all n).

Hence the question essentially asks us: If n is a positive integer and n^2 is divisible by 72, find n (or the largest value for n)

n^2 divisible by 72: This could be used as a test for elimination, and B,C,D,E all qualify as 'positive integers whose squares are divisible by 72'.

The largest valued choice, E, would be the most logical. What is the flaw in my reasoning above?

Thanks a ton,
vectorSpace
StaceyKoprince
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by StaceyKoprince Mon Jul 09, 2007 6:03 pm

The flaw lies in a crucial distinction between what n COULD be and what n MUST be (or MUST include). It's true that n COULD be divisible by 48, but it doesn't absolutely have to be.

Because of the wording, you can't rephrase this as "what is n" - or "what's the largest number n could be." You actually want the opposite - what's the smallest that n could be, because that will include the factors that MUST be part of n no matter what it actually is.
Stacey Koprince
Instructor
Director, Content & Curriculum
ManhattanPrep