Source: Official Guide Revised GRE 1st Ed. Part 6; Set 3; #14

8

Let S be the set of all positive

Let S be the set of all positive integers n such that n2 is a multiple of both 24 and 108. Which of the following integers are divisors of every integer n in S? Indicate all such integers. 12, 24, 36, 72, 15

4 Explanations

2

Hello, I am confused about the step where you convert 2^3 and 3^3 to 2^4 and 3^4. Are these numbers still multiples of 24 and 108 (as required in the question) and is there a rule for this (e.g. if you increase the exponents in a prime factorisation, the result will still be a multiple of the original numbers). Thank you.

Dec 3, 2017 • Comment

Sam Kinsman, Magoosh Tutor

Yes, (2^4 * 3^4) is a multiple of 24 and of 108. That is because, if we have the prime factorization of a number, and make the prime factorization greater by multiplying it by new numbers, the result will always be a multiple of the original number. Let me explain this using an example! :)

Let's use 100. The prime factorization of 100 is 2^2 * 5^2. If we make this number larger by multiplying by 2, we get 2^3 * 5^2. This is the same as having (2^2 * 5^2) * 2, which is the same as having (100) * 2. And 100 * 2 = 200, which is a multiple of 100.

So if you increase the exponents in a prime factorization, the result will still be a multiple of the original numbers.

I hope that helps! :)

Dec 15, 2017 • Reply

2

The explanation in the video does not appear to be complete as it only shows that all multiples of 36 are in the set S. To find a definitive answer, it is necessary to show that every member of S is a multiple of 36 too (i.e. the 2 sets are equal)

Sep 1, 2017 • Comment

Jonathan , Magoosh Tutor

Hi,
The explanation is complete. A number "n" can only be in S if "n^2" is multiple of 216; therefore, "n" MUST be a multiple of 36. So every member of S must be a multiple of 36, and all such positive multiples of 36 are in S. Therefore, because 36 is in S, the correct answers must be a factor of 36. Only 12 and 36 are factors of 36 among the answer choices. I hope that helps.

Sep 4, 2017 • Reply

1

Jessica Wilson

I still don't understand how you changed it to 2^4 3^4? Can you explain further?

Aug 8, 2017 • Comment

Sam Kinsman, Magoosh Tutor

Hi Jessica,

Good question!

We know that the smallest possible value for n^2 is 2^3 * 3^3. However, we know that n must be an integer. And if n^2 is 2^3 * 3^3, then n will not be an integer (you can test this by taking the square root of 2^3 * 3^3).

In general, if a number (let's call it x) has a square root that is an integer, it must be the case that that number (x) has a prime factorization where all the exponents are even.

Ok, so we know that the smallest possible value for n^2 is 2^3 * 3^3, but that doesn't work because n won't be an integer. So let's start with 2^3 * 3^3, and try to make that bigger so that n will be an integer. Well, to do this we need to make the exponents in the prime factorization even. So we convert this to 2^4 * 3^4. Here, we can take the square root, and it will be an integer (2^2 * 3^2). So if n^2 is 2^4 * 3^4, then n will be an integer.

If we start with 2^3 * 3^3, and make it bigger, the smallest possible number that is a perfect square is 2^4 * 3^4. So that's why we have to set n^2 = 2^4 * 3^4.

I hope this helps!

Sep 1, 2017 • Reply

10

Gravatar Chris Lele, Magoosh Tutor

Oct 8, 2012 • Comment

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