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Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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basic math

by omegan3 » Sun Jun 11, 2017 11:05 am
In one of the MGMAT books:
"If n is the product of 2,3, and a two-digit prime number, how many of its factors are greater than 6?"

In the answer they say:
"Because we have been asked for a concrete answer, we can infer that the answer will be the same
regardless of which 2-digit prime we pick."

Does anyone know why the answer doesn't depend on which 2-digit prime is picked?

Thanks.
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by [email protected] » Sun Jun 11, 2017 11:34 am
Hi omegan3,

There are a variety of different math 'rules' (patterns, Number Properties, etc.) that exist that most people don't realize are actually rules. Since GMAT questions are almost always based on a rule or pattern of some kind (and sometimes more than one), you can rely on the existence of those patterns and do the necessary work to get to the correct answer (although in DS questions, sometimes your work has to also focus on defining whether a combination of facts actually leads to a definitive pattern or not).

For this prompt, the rules involved are based on 'prime factorization', but even if you don't recognize that at first, you can still answer the question that is asked.

Let's start with the smallest two-digit prime number: 11.
(2)(3)(11) = 66
The factors of 66 are: 1 and 66, 2 and 33, 3 and 22, 6 and 11. Organized a different way, you can see them as...
1
2
3
11
(2)(3) = 6
(2)(11) = 22
(3)(11) = 33
(2)(3)(11) = 66

In this list, you'll see the number 1, each individual prime, the product of each "pair" of primes and the product of all 3 primes. THAT pattern will always occur regardless of what the 2-digit prime is. By extension, you should notice that - of the 6 factors - exactly four will always be greater than 6 (re: the 2-digit prime, 2(prime), 3(prime) and (2)(3)(prime)). If you don't 'see it' yet, then try another 2-digit prime (such as 13 or 17) and track the results - the answer to the question WILL be the same.

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by ceilidh.erickson » Sun Jun 11, 2017 2:37 pm
Broadly speaking, if a PS question asks for a concrete answer when the variable in question in unknowable (e.g. *some* 2-digit prime, but we'll never know which one), we have to infer that the answer would be the same no matter which value we pick, because otherwise there wouldn't be one (and only one) right answer to the question.

Here's another example:

PS #40 in OG2017:
If n is a prime number greater than 3, what is the remainder when n^2 is divided by 12 ?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 5
If there were different results for different values of n, then it would be impossible to answer the question.

So, we know that we can just pick any value for n that fits, and see what result we get:

n=5
n^2 = 25
25/12 give a remainder of 1.

The answer must be B.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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by omegan3 » Mon Jun 12, 2017 10:47 am
[email protected] wrote: ..
Rich
Hello Rich,

It's very nice to you to take the time to respond and help explain the reasoning.

You gave a very thorough explanation which I found helpful!
Thank you for that!
Last edited by omegan3 on Mon Jun 12, 2017 10:52 am, edited 1 time in total.
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by omegan3 » Mon Jun 12, 2017 10:51 am
ceilidh.erickson wrote: ..
Hi Ceilidh,

This seems like a useful way to think, a bit counter-intuitive for me, but I'll try to keep this reasoning in mind.

I also liked that you were able to come up with a similar example to explain you point!

Thank you.
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by ceilidh.erickson » Mon Jun 12, 2017 12:22 pm
omegan3 wrote:
ceilidh.erickson wrote: ..
Hi Ceilidh,

This seems like a useful way to think, a bit counter-intuitive for me, but I'll try to keep this reasoning in mind.

I also liked that you were able to come up with a similar example to explain you point!

Thank you.
My pleasure!
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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by Matt@VeritasPrep » Sun Jul 23, 2017 2:47 pm
Here's the simplest way, I think: call this prime p.

Since our number = 2 * 3 * p, it has the following factors:

1, 2, 3, p, 2*3, 2*p, 3*p, 2*3*p

Since p is a two digit number, everything here with p in it will be (at least) two digits. So the factors p, 2p, 3p, and 6p are all greater than 6, giving us FOUR such factors.
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by omegan3 » Mon Jul 24, 2017 6:04 am
Matt@VeritasPrep wrote:Here's the simplest way, I think: call this prime p.

Since our number = 2 * 3 * p, it has the following factors:

1, 2, 3, p, 2*3, 2*p, 3*p, 2*3*p

Since p is a two digit number, everything here with p in it will be (at least) two digits. So the factors p, 2p, 3p, and 6p are all greater than 6, giving us FOUR such factors.
Thank you dear for explaining, and sharing your knowledge with us.
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by Matt@VeritasPrep » Sun Aug 06, 2017 10:45 pm
omegan3 wrote:
Matt@VeritasPrep wrote:Here's the simplest way, I think: call this prime p.

Since our number = 2 * 3 * p, it has the following factors:

1, 2, 3, p, 2*3, 2*p, 3*p, 2*3*p

Since p is a two digit number, everything here with p in it will be (at least) two digits. So the factors p, 2p, 3p, and 6p are all greater than 6, giving us FOUR such factors.
Thank you dear for explaining, and sharing your knowledge with us.
Certainly! I don't know if I deserve being 'dear', but hey, I'm flattered.
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