Factorial

[vipulgoyal]

does the int K has factor p such that 1<p<k ??
1. k > 4 factorial
2. 13facto +2 <= K <= 13facto + 13

[Anju@Gurome]

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

(1) k > 4!
(2) 13! + 2 ≤ k ≤ 13! + 3

If an integer k have a factor p such that 1 < p < k, then k is a composite number, otherwise k is prime. Thus the question simply asks whether k is prime or composite.

Statement 1: k > 4!
k may be prime (29, 31 etc) or may be composite (25, 26 etc.)

Not sufficient.

Statement 2: (13! + 2) ≤ k ≤ (13! + 13)
Each possible value of k is composite integer. Take few for example,

• k = (13! + 2) = (13*12*11*...*5*4*3*2*1 + 2) = Some multiple of 2 + 2 = Multiple of 2
  k = (13! + 3) = (13*12*11*...*5*4*3*2*1 + 3) = Some multiple of 3 + 3 = Multiple of 3
  k = (13! + 4) = (13*12*11*...*5*4*3*2*1 + 4) = Some multiple of 4 + 4 = Multiple of 4
  Same for 5, 6, 7, 8, 9, 10, 11, 12 and 13

Sufficient

The correct answer is B.

[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).

[Mo2men]

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 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.

Dear Mitch,

I have a question about Statement 2, Why can't I cancel out 13! from both sides as 13! is added so it could be subtracted.

If I take simple example

3! + 3 ≤ K ≤ 3! + 7

6 + 3 ≤ K ≤ 6 + 7..... I can cancel out (subtract) 6 from both sides

Where did i get wrong?

Thanks in advance

[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 video with tips on rephrasing the target question: https://www.youtube.com/watch?v=wetqSGMdY7Y)

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

[GMATGuruNY]

Dear Mitch,

I have a question about Statement 2, Why can't I cancel out 13! from both sides as 13! is added so it could be subtracted.

If we subtract 13! from 13! + 2 ≤ k ≤ 13! + 13, we must apply this operation to EVERY PART of the inequality, as follows:
13! + 2 ≤ k ≤ 13! + 13

13! + 2 - 13! ≤ k - 13! ≤ 13! + 13 - 13!

2 ≤ k - 13! ≤ 13.

The algebra above is valid, but it does not seem to help us determine whether k is prime.

[ceilidh.erickson]

To add to what Mitch said, it doesn't help to subtract 13! from every part of the inequality, because the most helpful version of a 3-part inequality is to have a variable sandwiched between numbers:
e.g. 0 < x < 1
We can then read this as a range: "x is between 0 and 1."

It's not helpful in this case to say "the difference of k and 13! is between 2 and 13."

[ceilidh.erickson]

For further explanation of the original question, see here: https://www.beatthegmat.com/is-x-5-5-t276396.html#718640

And for more on translating tricky GMAT wording, see here: https://www.manhattanprep.com/gmat/blog ... uage-test/

[Jeff@TargetTestPrep]

does the int K has factor p such that 1<p<k ??
1. k > 4 factorial
2. 13facto +2 <= K <= 13facto + 13

We need to determine whether k has a factor p such that 1<p<k, or in other words, whether k is a prime number. If it is, then it doesn't have a factor between 1 and itself. If it isn't, then it does.

Statement One Alone:

k > 4!

Since there are prime numbers greater than 4! and composite (non-prime) numbers greater than 4!, statement one alone is not sufficient to answer the question.

Statement Two Alone:

13! + 2 ≤ k ≤ 13! + 13

13! will have a factor of any integers from 2 to 13 inclusive. For any number k is, in the range of integers from 13! + 2 to 13! + 13 inclusive, k will have a factor from 2 to 13 inclusive. In particular, if k = 13! + n where (2 ≤ n ≤ 13), k will have n as a factor. For example, if k = 13! + 5, then k will have a factor of 5, since 5 divides into 13! and 5. If k = 13! + 8, then k will have a factor of 8 and hence a factor of 2.

Thus, statement two is sufficient to answer the question.

Answer: B