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Re: If S is a sequence of consecutive multiples of 3 [#permalink]

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17 Jun 2014, 14:52

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]

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17 Jun 2014, 15:05

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]

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17 Jun 2014, 16:31

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]

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17 Jun 2014, 16:32

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]

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17 Jun 2014, 16:38

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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]

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18 Jun 2014, 02:27

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. _________________

If S is a sequence of consecutive multiples [#permalink]

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19 Aug 2014, 04:59

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.

1) statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked; 2) statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked; 3) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked; 4) EACH statement ALONE is sufficient to answer the question asked; 5) statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Re: If S is a sequence of consecutive multiples of 3 [#permalink]

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19 Aug 2014, 05:01

Expert's post

Saulius1 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.

1) statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked; 2) statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked; 3) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked; 4) EACH statement ALONE is sufficient to answer the question asked; 5) statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Merging similar topics. Please refer to the discussion above. _________________

Re: If S is a sequence of consecutive multiples of 3 [#permalink]

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15 Jul 2015, 22:16

Bunuel wrote:

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.

To make it more clear, S could be {120,123,126} which contains 1 multiple of 9 (126) or S could be {111,114,117,120,123,126} which contains 2 multiples of 9 (117 & 126)

So clearly insufficient.

Kudos please _________________

kudos please

gmatclubot

Re: If S is a sequence of consecutive multiples of 3
[#permalink]
15 Jul 2015, 22:16

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