Data Sufficiency

[gulshank1]

Answer with explanation pls.

[GMATGuruNY]

Does the integer k have a factor p such that 1 < p < k ?

(1) k > 4!
(2) 13! + 2 ≤ k ≤ 13! + 13

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).

Here's the mathematical reasoning behind statement 2:
13! + 2 = 2(13*12*11*10*9*8*7*6*5*4*3*1 + 1) = 2 * integer.
Thus, 2 is a factor of 13! + 2.

13! + 3 = 3(13*12*11*10*9*8*7*6*5*4*2*1 + 1) = 3 * integer.
Thus, 3 is a factor of 13! + 3.

13! + 4 = 4(13*12*11*10*9*8*7*6*5*3*2*1 + 1) = 4 * integer.
Thus, 4 is a factor of 13! + 4.

The same reasoning can be applied to every value that satisfies statement 2.
Thus, none of the values that satisfy statement 2 will be prime.

[Brent@GMATPrepNow]

Does the integer k have a factor p such that 1 < p < k ?

(1) k > 4!
(2) 13! + 2 ≤ k ≤ 13! + 13

Target question: Does the integer k have a factor p such that 1 < p < k ?

This question is a great candidate for rephrasing the target question. (We have a free video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100)

Let's look at a few cases to get a better idea of what the target question is asking.
- Try k = 6. Since 2 is a factor of 6, we can see that k DOES have a factor p such that 1<p<k.
- Try k = 10 Since 5 is a factor of 10, we can see that k DOES have a factor p such that 1<p<k.
- Try k = 16. Since 4 is a factor of 14, we can see that k DOES have a factor p such that 1<p<k.
- Try k = 5. Since 1 and 5 are the ONLY factors of 5, we can see that k does NOT have a factor p such that 1<p<k.
Aha, so if k is a prime number, then it CANNOT satisfy the condition of having a factor p such that 1 < p < k
In other words, the target question is really asking us whether k is a non-prime integer (aka a "composite integer")

REPHRASED target question: Is integer k a non-prime integer?

Statement 1: k > 4!
In other words, k > 24
This does not help us determine whether or not k is a non-prime integer? No.
Consider these two conflicting cases:
Case a: k = 25, in which case k is a non-prime integer
Case b: k = 29, in which case k is a prime integer
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 13! + 2 ≤ k ≤ 13! + 13
Let's examine a few possible values for k.

k = 13! + 2
= (13)(12)(11)....(5)(4)(3)(2)(1) + 2
= 2[(13)(12)(11)....(5)(4)(3)(1) + 1]
Since k is a multiple of 2, k is a non-prime integer

k = 13! + 3
= (13)(12)(11)....(5)(4)(3)(2)(1) + 3
= 3[(13)(12)(11)....(5)(4)(2)(1) + 1]
Since k is a multiple of 3, k is a non-prime integer

k = 13! + 4
= (13)(12)(11)....(5)(4)(3)(2)(1) + 4
= 4[(13)(12)(11)....(5)(3)(2)(1) + 1]
Since k is a multiple of 4, k is a non-prime integer

As you can see, this pattern can be repeated all the way up to k = 13! + 13. In EVERY case, k is a non-prime integer

Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = B

Cheers,
Brent

[gulshank1]

Thank You Gurus !! Once I rephrased, half of the question was solved, that was the catch
The link was very helpful.