Multiple

[mkamony]

Let S be the set of all positive integers n such that n^2 is a multiple of both 24 and 108. Which of the following are divisors of every integer in S?

select all that apply

A) 12
B) 24
C) 36
D) 72

[tommywallach]

Mkamony,

Please post using the format described in the forum header, allowing others to search for certain question explanations by the page # and question #. Otherwise we'll get reposted questions, which is inefficient for everyone.

This is a primes/divisibility question. You know that because it uses terms like "multiple" and "divisor", which are signals for p/d. Let's start by taking the factors of both numbers (standard operating procedure when dealing with primes and divisibility).

24 --> 2223
108 --> 33322

Squares always have an even number of each prime factor (because when you square a number, you double all the factors, which is like multiplying each one by two). So if our n^2 has three twos in it (which we know because 24 is a factor), it actually has to have four twos in it (because it's a square). Same goes for the threes we learn about in 108. So we know our n^2 has four twos in it and four threes. That means n itself must have two twos and two threes (half as many as n^2).

That means n will be a multiple of 2 * 2 * 3 * 3 = 36. 12 goes into 36, as does 36. 24 doesn't, nor does 72. So the answers should be 12 and 36.

This is a complex question, so let me know if you have any follow-up questions, or if you see an error.

Thanks!

-t