All the terms in Set S are integers. Five terms in S are eve

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.

B.

Correct.. the wording of the question is wrong.

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.

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.

Answer (B)

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

I am confused why using 2 to start the sequence does not rule out B

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

Evens - 5
Odds - 4

Evens not divisible by 3 = 4

6,7,8,9,10,11,12,13,14 gives us 3 evens not divisible by 3

2, 3, 4, 5, 6, 7, 8, 9, 10 doesn't satisfy all conditions. We need four multiples of 3 but it has only three (3, 6 and 9)

Same is the problem with this: 6,7,8,9,10,11,12,13,14

If you have consecutive integers and four of them are multiples of 3, two will be even and two will be odd. Hence, of the 5 even integers, only 2 will be multiples of 3.

the correct answer should be E , Because In question it's mentioned that All terms in s are integer.....(I can be positive or negative) Thus, Even with with only B option we can come to an answer but this answer is ambiguous because I can be +ve or -ve .

thus , I go with E.

For E you should demonstrate two cases giving two different answers. Kindly share your reasoning.

Even and odd integers are defined for negative numbers too so it won't matter.

\(?\,\,\, = \,\,\,x\,\,\, = \,\,\,5 - \left( {\# \,\,{\rm{multiples}}\,\,{\rm{of}}\,\,6} \right)\)




\(\left( 1 \right)\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,S = \left\{ {\,2\,,2 \cdot 3\,,2 \cdot {3^2},\,2 \cdot {3^{3\,}},2 \cdot {3^4}\,} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 5 - 4 = 1 \hfill \cr \\
\,{\rm{Take}}\,\,S = \left\{ {\,2\,,{2^2}\,,2 \cdot {3^2},\,2 \cdot {3^{3\,}},2 \cdot {3^4},3\,} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 5 - 3 = 2 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{INSUFF}}{\rm{.}}\)


\(\left( 2 \right)\,\,\,{\rm{different}}\,\,{\rm{classes}}\,\,{\rm{of}}\,\,{\rm{remainders}}\,\,{\rm{by}}\,\,3\,\,{\rm{and}}\,\,5\,\,\left( {{\rm{with}}\,\,5\,\,{\rm{even}}\,\,{\rm{numbers}}} \right)\,\,{\rm{:}}\,\,\,0,2,4\,\,{\rm{for}}\,\,{\rm{the}}\,\,{\rm{smaller}}\,\,{\rm{even}}\,\,{\rm{in}}\,\,{\rm{S}}\)

The idea is crucial: 0, 2 and 4 represent ALL possible scenarios for the first even integer belonging to S, even when negative integers are considered.
(We have presented the 6, 8 and 10 "next group" so that the "repetition of the cyclic behavior" becomes clear!)


\(\left. {\left\{ \matrix{\\
\,{\rm{0}}\,\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}} - 1} \right),1,3,5,7\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,9} \right)\,\, \to \,\,\,\,\,0,\,3,6,9\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 5 - 2 = 3 \hfill \cr \\
\,{\rm{2}}\,\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}}\,{\rm{ }}1} \right),3,5,7,9\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,11} \right)\,\, \to \,\,\,\,\,3,\,6,9\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,{\rm{not}}\,\,{\rm{viable}} \hfill \cr \\
\,{\rm{4}}\,\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}}\,{\rm{ }}3} \right),5,7,9,11\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,13} \right)\,\, \to \,\,\,\,\,3,\,6,9,12\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,? = 5 - 2 = 3 \hfill \cr \\
\,{\rm{6}}\,\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}}\,{\rm{ }}5} \right),7,9,11,13\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,15} \right)\,\, \to \,\,\,\,\,6,\,9,12,15\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,? = 5 - 2 = 3 \hfill \cr \\
\,{\rm{8}}\,\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}}\,{\rm{ }}7} \right),9,11,13,15\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,17} \right)\,\, \to \,\,\,\,\,9,\,12,15\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,{\rm{not}}\,\,{\rm{viable}} \hfill \cr \\
{\rm{10}}\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}}\,{\rm{ }}9} \right),11,13,15,17\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,19} \right)\,\, \to \,\,\,\,\,9,\,12,15,18\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,? = 5 - 2 = 3 \hfill \cr} \right.\,\,\,} \right\}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 3\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}{\rm{.}}\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.