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All the terms in Set S are integers. Five terms in S are eve [#permalink]
15 Jan 2013, 16:29

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E

Difficulty:

55% (hard)

Question Stats:

47% (02:45) correct
53% (02:55) wrong based on 78 sessions

All the terms in Set S are integers. Five terms in S are even, and four terms are multiples of 3. How many terms in S are even numbers that are not divisible by 3?

(1) The product of all the even terms in Set S is a multiple of 9. (2) The integers in S are consecutive.

Re: All the terms in Set S are integers. Five terms in S are eve [#permalink]
15 Jan 2013, 21:23

1) The product of all the even terms in Set S is a multiple of 9.

the smallest even multiple of 9 which should be a product of 5 even terms is-2x2x2x2x2x9, but there can other multiple of 9 too which should satisfy the basic criteria of the statement-like-2x2x2x2x4x9, etc-Not Sufficient

2) The integers in S are consecutive.

lets take an example:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

this will have 5 even terms but only 3 terms which are multiple of 3 and since 3 is part of only 1 even number-6, we cannot consider this

3, 4, 5, 6, 7, 8, 9, 10, 11, 12

now this will have 5 even terms and since 3 is now part of 2 even number-6 and 12, the product of even number will be a multiple of 9--hence sufficient

Re: All the terms in Set S are integers. Five terms in S are eve [#permalink]
05 May 2013, 04:58

I have a quick question.. Don't know someone has noticed it or not but the question never says that there are 9 elements in set S. It just says that there are 5 even numbers and 4 multiples of 3 which could overlap or even in one case there maybe other numbers besides these.

Re: All the terms in Set S are integers. Five terms in S are eve [#permalink]
05 May 2013, 05:56

Expert's post

Asishp wrote:

I have a quick question.. Don't know someone has noticed it or not but the question never says that there are 9 elements in set S. It just says that there are 5 even numbers and 4 multiples of 3 which could overlap or even in one case there maybe other numbers besides these.

Hence the answer should be E not B....

From F.S 1,we can have a series like 2,4,6,9,12,15,20. Here the even nos, not divisible by 3 are 3. Again, we could have another series like 4,6,12,18,24 and the even nos, not divisible by 3 is only 1. Insufficient.

From F.S 2, we know that the series is consecutive. For exactly 5 even integers, if our series starts with an even number, we need a total of 9 consecutive integers.

2k _ 2k+2 _ 2k+4 _ 2k+6 _ 2k+8 and we can have a maximum of three multiples of 3 as every 3 consecutive integers have one multiple of 3.

Thus, our series has to start with a odd number, and for placing 5 consecutive even integers, the series would be

_2k _ 2k+2 _ 2k+4 _ 2k+6 _ 2k+8 . Note that if the first integer is not a multiple of 3, we would still have 3 multiples of 3. Thus, the only way in which we can have 4 multiples is by having the first odd integer to be a multiple of 3. Thus, the number of even terms not divisible in such a case would be = 2k ,2k+4 and 2k+6. Sufficient.

Re: All the terms in Set S are integers. Five terms in S are eve [#permalink]
05 May 2013, 06:10

Asishp wrote:

I have a quick question.. Don't know someone has noticed it or not but the question never says that there are 9 elements in set S. It just says that there are 5 even numbers and 4 multiples of 3 which could overlap or even in one case there maybe other numbers besides these.

Re: All the terms in Set S are integers. Five terms in S are eve [#permalink]
05 May 2013, 06:18

1

This post received KUDOS

Expert's post

The terms do overlap (even and multiple of 3). Here is my explanation for B:

- If all terms are consecutive integers then 4 terms that are multiples of 3 follow each other too. If the first term is 3k (k is an integer) then the next one is 3(k+1) --> 3K+3. So, if 3k is odd, then 3k + 3 is even and if 3k is even then 3k+3 is odd.

There are only two combinations for those 4 terms: [even][odd][even][odd] (Example: 6, 9, 12, 15) or [odd][even][odd][even] (Example: 3, 6, 9, 12)

it means that there are always 2 even terms that are multiple of 3 and (5-2) terms that are not multiple of 3. Sufficient. _________________

Re: All the terms in Set S are integers. Five terms in S are eve [#permalink]
07 May 2013, 09:20

2

This post received KUDOS

Expert's post

superpus07 wrote:

All the terms in Set S are integers. Five terms in S are even, and four terms are multiples of 3. How many terms in S are even numbers that are not divisible by 3?

(1) The product of all the even terms in Set S is a multiple of 9. (2) The integers in S are consecutive.

So we have 5 even terms (so rest are all odd) and 4 multiples of 3. There may or may not be overlap in these two number types. All we can say right now is that there must be at least one even number which is not a multiple of 3 (since there are 5 even numbers and only 4 multiples of 3).

(1) The product of all the even terms in Set S is a multiple of 9.

The product of all even terms is a multiple of 3 doesn't tell us how many even numbers are divisible by 3. It is possible that only one even number has 9 as a factor and none of the other 4 even numbers have 3 as a factor. It is also possible that 4 even numbers have 3 as a factor and the product is divisible by 81 too. This statement doesn't imply that the product of all even numbers is divisible only by 9 and no higher power of 3. Not sufficient.

(2) The integers in S are consecutive. Since there are 5 even numbers, there must be either 4 or 5 or 6 odd numbers (s o that the numbers are consecutive) Out of 4 consecutive multiples of 3, two will be odd and 2 will be even. e.g. 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13. Every other multiple of 3 will be even and the rest will be odd. Hence, we have 2 even numbers which are multiples of 3 too. So, of the 5 even numbers, 2 must be multiples of 3 and 3 must not be multiples of 3 e.g. in our example, 6, 12 are multiples of 3 and 4, 8 and 10 will not be multiples of 3. Sufficient.

Re: All the terms in Set S are integers. Five terms in S are eve [#permalink]
07 May 2013, 09:41

walker wrote:

The terms do overlap (even and multiple of 3). Here is my explanation for B:

- If all terms are consecutive integers then 4 terms that are multiples of 3 follow each other too. If the first term is 3k (k is an integer) then the next one is 3(k+1) --> 3K+3. So, if 3k is odd, then 3k + 3 is even and if 3k is even then 3k+3 is odd.

There are only two combinations for those 4 terms: [even][odd][even][odd] (Example: 6, 9, 12, 15) or [odd][even][odd][even] (Example: 3, 6, 9, 12)

it means that there are always 2 even terms that are multiple of 3 and (5-2) terms that are not multiple of 3. Sufficient.

I really liked the way .. u explained Walker .Thanks !! ............................. _________________

If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

gmatclubot

Re: All the terms in Set S are integers. Five terms in S are eve
[#permalink]
07 May 2013, 09:41