Sequence of 3s

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Sequence of 3s

by kartikshah » Fri Jul 20, 2012 4:12 am
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.

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The OA is A and the solution provided is thus:

Stat. (1): Simulate the problem on a smaller scale of 3 terms: if set S includes 3 terms
{3, 6, 9}, then it has only one multiple of 9;
{6, 9, 12} - still only one multiple;
{9, 12, 15} - still only one multiple;
{12, 15, 18} - 9 is out, but 18 is in - still only one multiple.

This little exercise shows that every three multiples of 3 will include one multiple of 9. Therefore, if S includes 15 multiples of 3, then S will include 5 multiples of 9 - one for every three terms. Only 1 possible value for the number of multiples of 9 in S, so Stat. (1)->S->AD.

But I started by creating sets of 5 elements:
{3,6,9,12,15} = 1 multiple of 9
{9,12,15,18,21} = 2 multiples of 9

The above strategy didn't work for me so I selected C as my answer.

Can anyone point out the flaw? Is there a better way to solve similar problems?
Source: — Data Sufficiency |

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by Anurag@Gurome » Fri Jul 20, 2012 4:20 am
kartikshah wrote:But I started by creating sets of 5 elements:
With 5 terms the logic won't work as 5 is not a multiple of 3. But 15 is.

To be precise, the logic is : in each three consecutive multiples of 3, there will be only one multiple of 9.

Hence, in 15 consecutive multiples of 3, there will be 15/3 = 5 multiples of 9.
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