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Bunuel
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Does the integer k have a factor p such that 1 < p < k ? 12 Nov 2017, 09:28
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rudjlive
Hi Bunuel,

I have the same question as oryahalom.

Cant we subtract 13! from both sides of the inequality?

Thanks,
RD
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Are there only two parts in \(13! + 2 \leq k \leq 13!+13\)? No, there are three. If you subtract 13! you should subtract from all three parts and you'll end up with \(2 \leq k -13! \leq 13\), which gives you nothing.
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Does the integer k have a factor p such that 1 < p < k ? 19 Dec 2017, 07:46
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gmatcracker2010
Does the integer k have a factor p such that 1<p<k?


(1) k > 4!

(2) \(13! + 2 \leq k \leq 13!+13\)
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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
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Does the integer k have a factor p such that 1 < p < k ? 19 Aug 2019, 01:33
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Hi Bunuel,

If we go with 2nd option then we can have multiple factors

For ex: If we take k=13!+4 then apart from 4, it will have others factors also .

So we will be getting more than one values of P.

So is it correct to mark B as sufficient answer.

Please Help.

Thanks
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Does the integer k have a factor p such that 1 < p < k ? 19 Aug 2019, 01:45
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a12bansal
Hi Bunuel,

If we go with 2nd option then we can have multiple factors

For ex: If we take k=13!+4 then apart from 4, it will have others factors also .

So we will be getting more than one values of P.

So is it correct to mark B as sufficient answer.

Please Help.

Thanks
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This is not a value question. The question is NOT what is the value of p.

It's an YES/NO question: Does the integer k have a factor p such that 1<p<k? And from (2) we get a definite YES answer to that question.

Hope it's clear.
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Does the integer k have a factor p such that 1 < p < k ? 10 Dec 2019, 12:39
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the question is asking whether k has a factor that is greater than 1, but less than itself.
if you're good at these number property rephrasings, then you can realize that this question is equivalent to "is k non-prime?", which, in turn, because it's a data sufficiency problem (and therefore we don't care whether the answer is "yes" or "no", as long as there's an answer), is equivalent to "is k prime?".
but let's stick to the first question - "does k have a factor that's between 1 and k itself?" - because that's easier to interpret, and, ironically, is easier to think about (on this particular problem) than the prime issue.

--

key realization:
every one of the numbers 2, 3, 4, 5, ..., 12, 13 is a factor of 13!.

this should be clear when you think about the definition of a factorial: it's just the product of all the integers from 1 through 13. because all of those numbers are in the product, they're all factors (some of them several times over).

--

consider the lowest number allowed by statement 2: 13! + 2.
note that 2 goes into 13! (as shown above), and 2 also goes into 2. therefore, 2 is a factor of this sum (answer to question prompt = "yes").

consider the next number allowed by statement 2: 13! + 3.
note that 3 goes into 13! (as shown above), and 3 also goes into 3. therefore, 3 is a factor of this sum (answer to question prompt = "yes").

etc.
all the way to 13! + 13.
works the same way each time.
so the answer is "yes" every time --> sufficient.

--

Posted from my mobile device
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Does the integer k have a factor p such that 1 < p < k ? 07 May 2020, 05:50
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Bunuel - would it be correct to say that all factorials are composite numbers?
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Does the integer k have a factor p such that 1 < p < k ? 07 May 2020, 07:35
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Bunuel - would it be correct to say that all factorials are composite numbers?
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Yes, but except 0! = 1, 1! = 1, and 2! = 2.
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Does the integer k have a factor p such that 1 < p < k ? 04 Nov 2020, 16:19
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gmatcracker2010
Does the integer k have a factor p such that 1<p<k?


(1) k > 4!

(2) \(13! + 2 \leq k \leq 13!+13\)
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If the answer is NO -- if k does NOT have a factor p between 1 and k -- the implication is that k is PRIME.
Question stem, rephrased:
Is k prime?

Statement 1:
Clearly, many values greater than 4! will be prime, while most will not be prime.
Thus, the answer to the question stem can be YES or NO.
INSUFFICIENT.

Statement 2:
Since the GMAT cannot expect us to prove that any of the integers within the given range ARE prime, all of the integers within the given range must NOT be prime.
Thus, the answer to the question stem is NO.
SUFFICIENT.

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Does the integer k have a factor p such that 1 < p < k ? 05 Feb 2021, 09:23
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Bunuel could you please confirm can i use the below mentioned approach for the option 2?

(2) 13!+2≤k≤13!+13

Subtract 13! from each sides

2≤k-13!≤ 13
Now k must be 16!
so now there should be a 5*2 pair in k thus it is not a prime number .
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Does the integer k have a factor p such that 1 < p < k ? 05 Feb 2021, 10:08
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TAC470
Bunuel could you please confirm can i use the below mentioned approach for the option 2?

(2) 13!+2≤k≤13!+13

Subtract 13! from each sides

2≤k-13!≤ 13
Now k must be 16!
so now there should be a 5*2 pair in k thus it is not a prime number .
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No this is not correct. If k = 16!, then 16! - 13! is MUCH larger than 13.
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Does the integer k have a factor p such that 1 < p < k ? 05 Feb 2021, 11:36
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Bunuel
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Bunuel could you please confirm can i use the below mentioned approach for the option 2?

(2) 13!+2≤k≤13!+13

Subtract 13! from each sides

2≤k-13!≤ 13
Now k must be 16!
so now there should be a 5*2 pair in k thus it is not a prime number .

No this is not correct. If k = 16!, then 16! - 13! is MUCH larger than 13.
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thank you Bunuel and sorry for asking a dumb question just realised how silly my question was .
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Does the integer k have a factor p such that 1 < p < k ? 17 Jun 2021, 05:15
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If integer k has a factor p such that 1<p<k, then the integer is composite. If it doesn’t, then k is prime.
We can therefore rephrase the question as “Is k prime?”

From statement I alone, k > 4! i.e. k > 24.

This is not sufficient to find out if k is prime, because k could be any integer value greater than 24.
Statement I alone is insufficient. Answer options A and D can be eliminated. Possible answer options are B, C or E.

From statement II alone, 13! + 2 ≤ k ≤ 13! + 13.

It’s important to note that 13! will be divisible by all integers from 1 to 13. Therefore, regardless of the value of k, k will be divisible by at least one more integer other than 1 and itself.

k is NOT prime.
Statement II alone is sufficient to answer the question with a YES. Answer options C and E can be eliminated,

The correct answer option is B.

Hope that helps!
Aravind B T
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Does the integer k have a factor p such that 1 < p < k ? 30 Sep 2021, 17:29
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Help would be appreciated! (+Kudos if you had the same issue)
1) the question asked whether p is a factor of k, well with statement (B) if we assume p=17 (17 is less than 13!+2 so p=17 could be assumed) then p wont be a factor of k (because the largest prime k has is 13) , and if p=18 it would be a factor so we were not able to determine whether or not k has a factor of p (as it depends on the value of p),
2) the question hadnt mentioned whether p is an integer either (if p=2 then yes its a factor of k, if p = 1.5 then its not)
Many thanks!!
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Does the integer k have a factor p such that 1 < p < k ? 03 Jul 2023, 08:24
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Bunuel
Does the integer k have a factor p such that 1<p<k?

Question basically asks whether \(k\) is a prime number. If it is, then it won't have a factor \(p\) such that \(1<p<k\) (definition of a prime number).

(1) \(k>4!\) --> \(k\) is more than some number (\(4!=24\)). \(k\) may or may not be a prime. Not sufficient.

(2) \(13!+2\leq{k}\leq{13!+13}\) --> \(k\) can not be a prime. For instance if \(k=13!+8=8*(2*3*4*5*6*7*9*10*11*12*13+1)\), then \(k\) is a multiple of 8, so not a prime. Same for all other numbers in this range. So, \(k=13!+x\), where \(2\leq{x}\leq{13}\) will definitely be a multiple of \(x\) (as we would be able to factor out \(x\) out of \(13!+x\), the same way as we did for 8). Sufficient.

Answer: B.

Check similar question: https://gmatclub.com/forum/factor-facto ... 00670.html

Hope it's clear.
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Hey Bunuel, why did you multiple #2 by 8 if it was just adding 8?

For example if it said 5! + 11 it would be 120 +11 which would become a prime number but if it was 5! +10, it would not be a prime number
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Does the integer k have a factor p such that 1 < p < k ? 03 Jul 2023, 11:50
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Bunuel
Does the integer k have a factor p such that 1<p<k?

Question basically asks whether \(k\) is a prime number. If it is, then it won't have a factor \(p\) such that \(1<p<k\) (definition of a prime number).

(1) \(k>4!\) --> \(k\) is more than some number (\(4!=24\)). \(k\) may or may not be a prime. Not sufficient.

(2) \(13!+2\leq{k}\leq{13!+13}\) --> \(k\) can not be a prime. For instance if \(k=13!+8=8*(2*3*4*5*6*7*9*10*11*12*13+1)\), then \(k\) is a multiple of 8, so not a prime. Same for all other numbers in this range. So, \(k=13!+x\), where \(2\leq{x}\leq{13}\) will definitely be a multiple of \(x\) (as we would be able to factor out \(x\) out of \(13!+x\), the same way as we did for 8). Sufficient.

Answer: B.

Check similar question: https://gmatclub.com/forum/factor-facto ... 00670.html

Hope it's clear.

Hey Bunuel, why did you multiple #2 by 8 if it was just adding 8?

For example if it said 5! + 11 it would be 120 +11 which would become a prime number but if it was 5! +10, it would not be a prime number
Show more

We did not multiply by 8, we factored out 8 from 13! + 8 = 2*3*4*5*6*7*8*9*10*11*12*13 + 8 to get 8*(2*3*4*5*6*7*9*10*11*12*13+1).

Hope it helps.
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Does the integer k have a factor p such that 1 < p < k ? 07 Jul 2023, 15:55
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Thanks very much for the clarification Bunuel!
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Does the integer k have a factor p such that 1 < p < k ? 23 Feb 2025, 07:47
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Hey Bunuel

Had the Statement 2 said '13!+2 ≤ k ≤ 13!+17', then what would the solution be?
Want to gain more clarity on the concept of factoring out. Thanks!
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Does the integer k have a factor p such that 1 < p < k ? 16 Jun 2025, 04:45
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We Can say...Is there any number between 1 and K (exclusive) that is a factor of K.......
For any non prime...Yes several Numbers could be between 1 to K that are factors of K
For any Prime...No ..So Such number could exist....

Note: We don't want to know what is P...but whether P can exist or Not...


1....k>4!....Not Sufficient....... K >4! Could be anything...5! - Means a prime..Means there exist a number less than 5! that could be its factor...Could also be 4! + 5 ....29...No Number less than 29 Exists that is a factor of 29.....

2. K is integer between 13! + 2...and 13! + 13 .....All are non prime.....

So Does the integer k have a factor p such that 1<p<k? With Statement 2 YES.....all of them have atleast one factor which is less than k......

The point here is ...whether there exists ANY Factor that is less than K....There may be several numbers which are not factors of K..but as long as we have just 1 factor..we can be sure that its a non prime...We don't care what is the value of P....

Question essentially is...Is there any factor whose value is less than K and more than 1....Means if it is prime or not...



Do correct me if i am wrong anywhere..Thanks






gmatcracker2010
Does the integer k have a factor p such that 1<p<k?


(1) k > 4!

(2) \(13! + 2 \leq k \leq 13!+13\)
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