hey i think the answer is E

I'd like to see the explanation because I get E.

If there are exactly 3 multiples of 6 in a sequence of 13 consecutive integers, then that sequence must start and end with multiples of 6.

1) The sequence cannot contain 25, 35, 55, 65, 85 or 95. Not sufficient because this still allows, for example, the sequences 6...18 and 66...78, which have different numbers of primes.
2) The sequence must contain one of 7, 35, 49, 77 or 91. Not sufficient because this allows, for example, the sequences 6...18 and 30...42, which have different numbers of primes.

With both statements combined, I looked at sequences that included numbers specified in 2) and then checked to see if they included numbers specified in 1).
7 - 6...18
35 - can't be included because of 1).
49 - 42...54
77 - 66...78, 72...84
91 - can't be included without including 95 or going over 100.

So there are 4 sequences that have 3 sixes, contain two multiples of 5 that are also multiples of 2 or 3, and exactly one multiple of 7 that is not also a multiple of 2 or 3. 6...18 has 4 primes (7,11,13,17) while the other three sequences have 3.

does PR cat has such good questions ??

answer must be E.

even after adding both statement sequence can be :

6-18(7 is not a multiple of 2 or 3.)
26-38(35 is not a multiple of 2 or 3.) and so on

hence answer is E

@shoot4greatness: Can you please provide the explanation why OA is C??

This question was discussed in the thread linked below. Ian Stewart agrees that, as written, the answer should be E. He guesses that the question meant to state that all integers are positive, and that all integers are larger than 10. He then gives an interesting way to solve the newly constrained question.

https://www.beatthegmat.com/please-help-me-t17777.html