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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

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).

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.

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

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15 Jul 2015, 22:16

1

This post received KUDOS

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)

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

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03 Aug 2016, 07:28

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

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05 Oct 2016, 08:11

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.

Νice one. Picked A because I failed to consider 0 as a multiple of 9. Nice explanation
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Re: If S is a sequence of consecutive multiples of 3 [#permalink]

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09 Dec 2016, 04:31

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.

Hi Bunuel! Thanks for the great explanation!

When I tried to solve the problem, for statement 1 I also broke down the set of 15 numbers into a smaller set. However, I chose a set of 5 to test, which obviously gave me the wrong answer. How do we know that a set of 3 is appropriate to test?

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

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18 May 2017, 05:19

Hi, After spending a good amount of time on this problem to find an easy way to solve this problem I found a pattern. At first I took an example of the set of 15 numbers which are multiples of 3: 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 Then I started to find a pattern here that after every two numbers there is a multiple of 9 in the series. I checked this for 12 through 54 and -9 through 33. In both the series we are getting multiples of 9 after every 2 numbers. Thus the pattern looks something like below for 3 through 9 series: 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 I have taken 0 for the numbers in the series which are not divisible by 9 and 1 for the numbers which are divisible by 9. For two other set of 15 numbers which are multiples of 3, 12 through 54 and -9 through 33, we will end with the same number of ones, that is, multiples of 9. Also on close observing I found that by the distance of the first multiple of a given number from zero will give us a pattern. It works for other numbers too. Consider a case where we have to find the number of multiples of 6 from a set of multiples of 3 and if we know the number of terms is 15 we can get the pattern as below: 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 there are 7 multiples of 6 in a set of 15 multiples of 3 If we consider the same series: 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 As zero is a multiple of all the numbers and there is only one number in between 0 and 6 (first multiple of 6), we can safely say that the distance between two consecutive multiples of 6 in a set of multiples of 3. Based on that we can write down the 0101… series and find out the number of multiples of 6 or 9 or what so ever asked in the question. And I guess we can apply this to other problems as well. Even I am trying to find the best approach to solve problem like you all , please let me know if I am going wrong somewhere with this approach or if this approach itself is wrong. Because I have not checked this for too many numbers.
_________________

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

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13 Sep 2017, 01:20

hideyoshi wrote:

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

Thanks, this cleared it up for me! I kept wondering how 126 wouldn't be enough (I counted down till first term was 3) but that's not what it implies.

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