gmatcracker2010 wrote:
Does the integer k have a factor p such that 1<p<k?
(1) k > 4!
(2) \(13! + 2 \leq k \leq 13!+13\)
For K >4 ! (=24)
k can be prime or composite. k = 29, 31 (which are prime) and k = 25, 26, 27 etc which are composite.
So the answer in case K is prime is No. But in case k is composite, the answer is Yes....as there will be factors which lie between the number and 1.
For 13! + 2 <= k <=13! +13, k can be anything from 13!+2, 13!+3, 13! + 4, .......until 13!+13. (in all 12 numbers).
In all these possibilities, a factor can be found which is greater than 1 and less than k.
For example. in 13! + 2, 2 can be taken as common as 13! will certainly have 2. so 2 is a factor
in 13! + 7, 7 can be taken common from 13! and 7. So 7 is a factor.
So in all these scenarios, we are sure to find a factor.
Had this been 13! + 1, it would have been difficult to find if there is any factor for this or not.
Let me know if you still have any doubt.