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A) k > 4! means k > 24 let K be 25 which has factors 1, 5 and 25
1 < 5 < 25
But 29 has factors 1 and 29 and does not satisfy the given condition
So A is insufficient.
B) 13! has all the integer mutiples from 1 and 13
13!+2 is divisible by 2
13!+3 is divisible by 3 and so on
13!+12 is divisible by 12
For all of these 1 < n(from 2 to 12) < 13!+n

I wont be surprised if the answer is D. The stem of the problem is confusing. But I am pretty sure that B is the answer.

1) k>4! involves infinite numbers. This set holds primes and non primes.
Only primes do not satisfy the condition 1<p<k
2) Now 13! = 13*12*11*10*9*8*7*6*5*4*3*2*1 = X
This huge number is divisible by any integer between 1 and 13.
The second condition says that the number K is in between
13!+2 and 13!+13. So what are the possible numbers

13*12*11*10*9*8*7*6*5*4*3*2*1 + 3 divisible by p=3 1<3<(X+3)
13*12*11*10*9*8*7*6*5*4*3*2*1 + 4 divisible by p=4 1<4<(X+4)
----
----
13*12*11*10*9*8*7*6*5*4*3*2*1 + 12 divisible by p=3 1<12<(X+12)

So any number you take it has a factor > 1 and < that number.
Hence B is sufficient

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