ela07mjt wrote:Does the integer k have a factor p such that 1 < p < k ?
(1) k > 4!
(2) 13! + 2 ≤ k ≤ 13! + 13
The value in red had been posted incorrectly.
I've amended statement 2 to reflect the problem's original intent.
If k has NO FACTOR greater than 1 and less than k, then the only factors of k are 1 AND K ITSELF, implying that k is PRIME.
Question rephrased: Is k prime?
Statement 1: k > 4!
4! = 4*3*2 = 24.
If k = 29, then k is prime.
If k = 25, then k is not prime.
INSUFFICIENT.
Statement 2: 13! + 2 ≤ k ≤ 13! + 13
Apply a bit of REASON.
The values here are HUGE.
There is no way for us to prove that a huge number is prime.
Thus, every value of k that satisfies statement 2 must be NON-PRIME, since it would be impossible for us to prove that any of these values ARE prime.
SUFFICIENT.
The correct answer is
B.
A useful take-away:
If a DS problem asks whether a HUGE NUMBER is prime, the answer almost certainly will be NO, since there is no way for us to prove that a huge number actually IS prime (unless we're told rather directly that it has no factors other than 1 and itself).
Last edited by
GMATGuruNY on Mon Apr 22, 2013 2:50 am, edited 5 times in total.
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