GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 05 Jul 2020, 03:11

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 64947
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

12 Nov 2017, 08:28
2
rudjlive wrote:
Hi Bunuel,

I have the same question as oryahalom.

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

Thanks,
RD

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.
_________________
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 11028
Location: United States (CA)
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

19 Dec 2017, 06:46
gmatcracker2010 wrote:
Does the integer k have a factor p such that 1<p<k?

(1) k > 4!

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

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.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
225 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

Intern
Joined: 13 Dec 2018
Posts: 31
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

19 Aug 2019, 00:33
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.

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 64947
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

19 Aug 2019, 00:45
1
a12bansal wrote:
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.

Thanks

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.
_________________
Manager
Status: wake up with a purpose
Joined: 24 Feb 2017
Posts: 195
Concentration: Accounting, Entrepreneurship
Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

10 Dec 2019, 11:39
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
_________________
If people are NOT laughing at your GOALS, your goals are SMALL.
Stanford School Moderator
Joined: 11 Jun 2019
Posts: 104
Location: India
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

07 May 2020, 04:50
Bunuel - would it be correct to say that all factorials are composite numbers?
Math Expert
Joined: 02 Sep 2009
Posts: 64947
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

07 May 2020, 06:35
1
anindhya25 wrote:
Bunuel - would it be correct to say that all factorials are composite numbers?

Yes, but except 0! = 1, 1! = 1, and 2! = 2.
_________________
Intern
Joined: 07 May 2020
Posts: 20
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

07 Jun 2020, 18:07
Hi Bunuel - can you confirm whether or not it is incorrect to simply substitute x for y? 2 - 3/2 = -1/2
Math Expert
Joined: 02 Sep 2009
Posts: 64947
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

07 Jun 2020, 23:29
LaurenGol wrote:
Hi Bunuel - can you confirm whether or not it is incorrect to simply substitute x for y? 2 - 3/2 = -1/2

I think you are asking about some other question. Also, 2 - 3/2 = 1/2, not -1/2.
_________________
Re: Does the integer k have a factor p such that 1 < p < k ?   [#permalink] 07 Jun 2020, 23:29

Go to page   Previous    1   2   [ 29 posts ]