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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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 \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), 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.

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

Question basically asks is \(k\) 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 \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), 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 \(13!+x\)). Sufficient.

Does the integer k have a factor p such that 1<p<k? [#permalink]
03 Nov 2010, 14:51

2

This post received KUDOS

Expert's post

metallicafan wrote:

Bunuel wrote:

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

Question basically asks is \(k\) 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 \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), 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 \(13!+x\)). Sufficient.

Thanks Bunuel. I have a question, what would happen if the second statement said this?

(2) \(13!+1\leq{k}\leq{13!+13}\)

I have seen some similar problems, in which they add 1. Could you post some links about it?

Thanks!

So basically we should determine whether \(13!+1\) is a prime number (as shown above all other possible values of k are not are not prime), which cannot be done without a computer. There are some particular values of \(k=n!+1\) for which we can say whether it's a prime or not with help of Wilson's theorem, but again it's out of the scope of GMAT.

By the way: \(13!+1\) is not a prime number, it has two distinct prime factors: \(13!+1=83*75,024,347=6,227,020,801\), so the answer still will be B. _________________

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 \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), 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 \(13!+x\)). Sufficient.

Sorry for the question, but I am just not understanding how you get 8*(2*4*5*6*7*9*10*11*12*13+1) from 13!+8 then how do you know that k is a multiple of 8...I obviously have some deficiencies when it comes to number properties. So then how is k never a prime between those parameters? I would think (1*2*3*4*5*6*7*8*9*10*11*12*13) +8 ...

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 \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), 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 \(13!+x\)). Sufficient.

Sorry for the question, but I am just not understanding how you get 8*(2*4*5*6*7*9*10*11*12*13+1) from 13!+8 then how do you know that k is a multiple of 8...I obviously have some deficiencies when it comes to number properties. So then how is k never a prime between those parameters? I would think (1*2*3*4*5*6*7*8*9*10*11*12*13) +8 ...

k=13!+8 means k=2*3*4*5*6*7*8*9*10*11*12*13+8, now factor out 8: k=8*(2*4*5*6*7*9*10*11*12*13+1) --> k is a multiple of 8 as k=8*something --> as k is a multiple of 8 it can not be a prime number. You can have any number from 2 to 13 inclusive instead of 8, and you'll be able to factor out this number the same way as you did with 8, so any number of a type 13!+x, wher x is from 2 to 13 inclusive will be a multiple of x, thus not a prime number.

Check the link in my first post for similar problem. Also check Number Theory chapter of Math Book: math-number-theory-88376.html

Answer can't be E, it has to be B. I don't have OG with me, but I need to check.

1. By itself not sufficient. K could be prime. 2. 13! + 2 <= K <= 13! + 13 ensures that you have ever number from 2 - 13 dividing k. That is: 13! + 2 -> div by 2 13! + 3 -> div by 3 ... so on till 13.

So, answer should be B. _________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Does the integer k have a factor p such that 1 < p < k ? [#permalink]
29 Jan 2012, 15:21

28

This post received KUDOS

Expert's post

16

This post was BOOKMARKED

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.

Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
14 Feb 2012, 13:27

Hi bunuel I don't understand the definition of a prime ( 1<p<k ). I know what a prime is but this drives me nuts. For an example I take 10 for k. I have factors 2x5 that are greater than 1 and smaller than 10.

Or is it asking for one factor? I see "a factor" in the question, which is singular. Then I understand this. It is really tricky.

Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
14 Feb 2012, 15:15

3

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

M3tm4n wrote:

Hi bunuel I don't understand the definition of a prime ( 1<p<k ). I know what a prime is but this drives me nuts. For an example I take 10 for k. I have factors 2x5 that are greater than 1 and smaller than 10.

Or is it asking for one factor? I see "a factor" in the question, which is singular. Then I understand this. It is really tricky.

Question asks whether some number k has a factor p which is more than 1 but less than k. For example if k=10 then the answer is yes, since both 2 and 5 are factors of 10 and are more than 1 and less than 10. But if for example k=7=prime then the answer is no, since 7 has no factor which is more than 1 and less than 7.

Now, look at the definition of a prime number: a prime number is a positive integer with exactly two factors: 1 and itself. So, we can say that the questions asks whether k is a prime number, because if it is then it won't have a factor which is more than 1 and less than k.

Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
18 Nov 2012, 02:18

Bunuel wrote:

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 \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), 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.

Hi Bunuel, I have a small query here- Had the question been : 4!+2<k<4!+6, if I follow the approach explained above I may conclude that k is not a prime no. reason being all the factors of 4! are the factors of 4!+?. Since its easy to calculate here the values between 4!+2 and 4!+6, I already know that it includes 29 which is a prime no., where as in case of 13!+2<k<13!+13, it is not easy to calculate 13!, we may be missing some value which is a prime no. How do we make sure that we are not missing anything? (Assuming we can not cross check by calculating the values because of the time constraint at the exam time)

Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
18 Nov 2012, 04:01

Expert's post

mneeti wrote:

Bunuel wrote:

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 \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), 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.

Hi Bunuel, I have a small query here- Had the question been : 4!+2<k<4!+6, if I follow the approach explained above I may conclude that k is not a prime no. reason being all the factors of 4! are the factors of 4!+?. Since its easy to calculate here the values between 4!+2 and 4!+6, I already know that it includes 29 which is a prime no., where as in case of 13!+2<k<13!+13, it is not easy to calculate 13!, we may be missing some value which is a prime no. How do we make sure that we are not missing anything? (Assuming we can not cross check by calculating the values because of the time constraint at the exam time)

If \(13!+2\leq{k}\leq{13!+13}\), then we can factor out \(x\) out of \(13!+x\) (where \(2\leq{x}\leq{13}\)), which means that k is not a prime.

But you cannot apply the same logic with 4!+2<k<4!+6, since if k=4!+5, then you cannot factor out 5 out of it. _________________

Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
27 Jan 2013, 20:25

So basically in the second statement we can factor out a number from 2-13 but if its higher than that we can't say for sure it would be a prime or not? _________________

The question is basically asking whether K is prime or not. If it is prime then P can be 1 or k else it will follow 1<p<k.

Statement 1. k>4! k>24; let's say k=25, then p=5; we get a Yes to the question. if k=29; then p=1 or 29; we get a No to the question. Thus, insufficient.

Statement 2: 13!+2<=k<=13!+13 k can take any value from 13!+2 to 13!+13. All the values in this range has one thing in common. They all can have one common value: for e.g. 13!+2= 2[(13*12*...except 2)+1] 13!+10=10[(13*12*...(except 10)+1] No matter whether the value in the bracket is prime or not; the common value will make it non prime and will always give a yes to the question. Thus, sufficient.

Essentially this is about prime and non-prime. Option 1: k>4! Now assume k=5. then is there any factor of 5 which lies between 1 and 5? NO. Take k=6. then there is 2,3 which lies between 1 and 6. So 1 alone insufficient.

Option 2: 13!+2<=k<=13!+13 Take say 13!+3. now it can be rewritten as 3*{(13.12.11.....4.2.1)+1} So there is 3 as factor between 1 and k. Similarly you can take the common factor out of all nos. in option 2. Therefore option 2 is sufficient! Add Kudos if this helps!

gmatclubot

Re: Does the integer k have a factor p such that 1 < p < k ?
[#permalink]
29 May 2013, 22:17

Low GPA MBA Acceptance Rate Analysis Many applicants worry about applying to business school if they have a low GPA. I analyzed the low GPA MBA acceptance rate at...

UNC MBA Acceptance Rate Analysis Kenan-Flagler is University of North Carolina’s business school. UNC has five programs including a full-time MBA, various executive MBAs and an online MBA...

To hop from speaker to speaker, to debate, to drink, to dinner, to a show in one night would not be possible in most places, according to MBA blogger...

Most top business schools breed their students for a career in consulting or financial services (which is slowly being displaced by high tech and entrepreneurial opportunities). Entry into...