Data Sufficiency

In a sequence of 13 consecutive integers, all of which are less than 100, there are exactly 3 multiples of 6. How many integers in the sequence are prime?

(1) Both of the multiples of 5 in the sequence are also multiples of either 2 or 3.

(2) Only one of the two multiples of 7 in the sequence is not also a multiple of 2 or 3.

[spoiler]OA C

But I get the answer as E.
I get 2 solutions :
6,7,8,9,10,11,12,13,14,15,16,17,18 has 4 primes and
42,43,44,45,46,47,48,49,50,51,52,53,54 has 3 primes

Original explanation:
This one is a group problem in disguise. First, there's a lot of information buried in the question. If there are exactly three multiples of 6 in the sequence, then the sequence must start with a multiple of 6. So, there will always be 7 multiples of 2 and 5 multiples of 3 in the sequence. Those are two of the groups in the group formula. The total is, of course, 13. We'll also need to subtract the numbers that are both multiples of 2 and 3 (aka the multiples of 6) from the total. That means that the statements need to tell us about the multiples of 5 and 7. Everything else in the sequence (neither in the group formula) will be prime. Statement (1) tells us that we don't need to count either of the multiples of 5 since they have already been counted but it tells us nothing about the multiples of 7 in the sequence. Hence, BCE. Statement (2) tells us to count one of the multiples of 7 but tells us nothing about the multiples of 5. Cross off B. Finally, when we put the statements together we know that 13 = 7(the multiples of 2) + 5 (the multiples of 3) + 2 (the multiples of 5) + 2 (the multiples of 7) − 3 (the multiples of 6) − 2 (the multiples of 5 and 2 or 3) − 1 (the multiples of 7 and 2 or 3) + Primes. Hence, there are 3 primes in the sequence. The correct answer is C.[/spoiler]