From statement A you cannot say anything. K could be 25 and still satisfy the condition 1<5<25

From the 2nd statement: You know that 13! ends in a 0
If you add 2 to this, the number is divisible by 2 thus 1<2<13!+2
Similarly for 3,4,5,6,7,8,9,10,11,12 and 13

So 2 is sufficient.

For any of the even numbers, you know that there exists a number
1<2<k for all even k between 13!+2 and 13!+13.

However, how do you know that none of the odd numbers between 13!2 and 13!+13 are not prime?

m_blooms wrote:Does the integer k have a factor p such that 1<p<k

(1) "k>4!"

(2) "13!+2 <= k <= 13!+13"



I know its asking something about prime numbers...HELP PLEASE?

The question is really asking "is k prime?"

If k is prime, then there will be no factor of k between 1 and k.

If k is not prime, then there will be a factor of k between 1 and k.

The only exception is that if k is less than or equal to 1, of course there will be no factors between 1 and k.

(1) k > 4

Well, there are lots of primes and non-primes greater than 4: insufficient.

(2) 13! + 2 <= k <= 13! + 13

There will be no primes in this range, so this statement is sufficient.

Here's how we know:

Every number from 1 to 13 is a factor of 13!.

Since 2 is a factor of 13!, 2 will also be a factor of 13! + 2
Since 3 is a factor of 13!, 3 will also be a factor of 13! + 3
Since 4 is a factor of 13!, 4 will also be a factor of 13! + 4
.
.
.
Since 13 is a factor of 13!, 13 will also be a factor of 13! + 13

(2) is sufficient and (1) isn't: choose (b).