Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: If S is a sequence of consecutive multiples of 3 [#permalink]
17 Jun 2014, 13:52

Expert's post

If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?

(1) There are 15 terms in S.

In every set of 3 consecutive multiples of 3, there will be one multiple of 9: {-9, -6, -3} {-6, -3, 0} {-3, 0, 3} {0, 3, 6} {3, 6, 9} ...

15 consecutive multiples of 3 could be break down into 5 different such sets, each of which will contain 1 multiple of 9. Therefore, there are 5 multiples of 9 in the set. Sufficient.

(2) The greatest term of S is 126. Clearly insufficient.

Re: If S is a sequence of consecutive multiples of 3 [#permalink]
17 Jun 2014, 14:05

ajithkumar wrote:

If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?

(1) There are 15 terms in S.

(2) The greatest term of S is 126.

I don't agree with the official answer. Any thoughts on this question?

Indeed the correct answer is (A) because whatever sequence with 15 terms you pick you will always have 5 multiples of 9, whether you start the sequence with a multiple of 9 or not (there's a pattern: every 3 numbers there's a multiple of 9 and in a 15 terms there're 5 sets of 3 terms ) Hope that helps

Re: If S is a sequence of consecutive multiples of 3 [#permalink]
17 Jun 2014, 15:31

Bunuel wrote:

If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?

(1) There are 15 terms in S.

In every set of 3 consecutive multiples of 3, there will be one multiple of 9: {-9, -6, -3} {-6, -3, 0} {-3, 0, 3} {0, 3, 6} {3, 6, 9} ...

15 consecutive multiples of 3 could be break down into 5 different such sets, each of which will contain 1 multiple of 9. Therefore, there are 5 multiples of 9 in the set. Sufficient.

(2) The greatest term of S is 126. Clearly insufficient.

Answer: A.

{-6, -3, 0} {-3,0,3} in these sets there are no multiples of 9..

Re: If S is a sequence of consecutive multiples of 3 [#permalink]
17 Jun 2014, 15:32

clipea12 wrote:

ajithkumar wrote:

If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?

(1) There are 15 terms in S.

(2) The greatest term of S is 126.

I don't agree with the official answer. Any thoughts on this question?

Indeed the correct answer is (A) because whatever sequence with 15 terms you pick you will always have 5 multiples of 9, whether you start the sequence with a multiple of 9 or not (there's a pattern: every 3 numbers there's a multiple of 9 and in a 15 terms there're 5 sets of 3 terms ) Hope that helps

you are not considering 0, and negative multiples of 3. So the answer should be C

Re: If S is a sequence of consecutive multiples of 3 [#permalink]
17 Jun 2014, 15:38

1

This post received KUDOS

Expert's post

ajithkumar wrote:

Bunuel wrote:

If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?

(1) There are 15 terms in S.

In every set of 3 consecutive multiples of 3, there will be one multiple of 9: {-9, -6, -3} {-6, -3, 0} {-3, 0, 3} {0, 3, 6} {3, 6, 9} ...

15 consecutive multiples of 3 could be break down into 5 different such sets, each of which will contain 1 multiple of 9. Therefore, there are 5 multiples of 9 in the set. Sufficient.

(2) The greatest term of S is 126. Clearly insufficient.

Answer: A.

{-6, -3, 0} {-3,0,3} in these sets there are no multiples of 9..

So the answer is C isn't it?

0 is divisible by EVERY integer except 0 itself, (or, which is the same, zero is a multiple of every integer except zero itself).

Re: If S is a sequence of consecutive multiples of 3 [#permalink]
18 Jun 2014, 01:27

Expert's post

pretzel wrote:

For the second statement, 126 terms can be divided into 42 sets isn't it?

The second statement does not say that there are 126 terms in the set. It says that the greatest term of S is 126, there can be any number of elements.