All the terms in Set \(S\) are integers. Five terms in \(S\)

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Source: Manhattan Prep

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.

The OA is B
Source: — Data Sufficiency |

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by Jay@ManhattanReview » Wed Aug 07, 2019 12:38 am

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BTGmoderatorLU wrote:Source: Manhattan Prep

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.

The OA is B
Let's take each statement one by one.

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

If one of the five integers is a multiple of 9, then others are not; thus, the count is 4; however, if only two of the five integers are multiples of 3, then others are not; thus, the count is 3. No unique answer. Insufficient.

2) The integers in \(S\) are consecutive.

Given that the integers in \(S\) are consecutive, and five terms are even, the number of odd terms could be 4/5/6. So, there would be 9, 10 or 11 terms.

We know that four terms are multiples of 3. The even terms that are multiples of 3 must be multiples of 6. Since the multiples of 6 are spaced at an interval of 6, there must be at least two even multiples of 6 (6 < minimum number of terms, 9). However, to have three multiples of 6, we must have at least 13 terms. Since the maximum number of terms is 11 (11 < 13), this is not the possibility. So, there are two even numbers that are divisible by 3; thus, there are three even numbers that are not divisible by 3. Sufficient.

The correct answer: B

Hope this helps!

-Jay
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