Join us for MBA Spotlight – The Top 20 MBA Fair      Schedule of Events | Register

 It is currently 06 Jun 2020, 06:18 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Does the integer k have a factor p such that 1 < p < k ?

Author Message
TAGS:

### Hide Tags

Manager  Joined: 13 May 2010
Posts: 81
Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

14
178 00:00

Difficulty:   55% (hard)

Question Stats: 60% (01:36) correct 40% (01:58) wrong based on 1801 sessions

### HideShow timer Statistics

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

(1) k > 4!

(2) $$13! + 2 \leq k \leq 13!+13$$
Math Expert V
Joined: 02 Sep 2009
Posts: 64318
Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

63
2
86
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.

Check similar question: http://gmatclub.com/forum/factor-factorials-100670.html

Hope it's clear.
_________________
##### General Discussion
Retired Moderator Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 873
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

2
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.

Check similar question: factor-factorials-100670.html

Hope it's clear.

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!
_________________
"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

GMAT Club Premium Membership - big benefits and savings
Math Expert V
Joined: 02 Sep 2009
Posts: 64318
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

3
1
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.

Check similar question: factor-factorials-100670.html

Hope it's clear.

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.
_________________
Intern  Joined: 12 Nov 2010
Posts: 18
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

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 $$13!+x$$). Sufficient.

Check similar question: factor-factorials-100670.html

Hope it's clear.

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 ...
Math Expert V
Joined: 02 Sep 2009
Posts: 64318
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

6
2
ChenggongMAS 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 $$13!+x$$). Sufficient.

Check similar question: factor-factorials-100670.html

Hope it's clear.

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

Hope it's clear.
_________________
Intern  Joined: 06 Nov 2011
Posts: 33
Location: Germany
Concentration: Entrepreneurship, General Management
GMAT Date: 03-10-2012
GPA: 3
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

1
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. Posted from GMAT ToolKit
Math Expert V
Joined: 02 Sep 2009
Posts: 64318
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

11
5
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. Posted from GMAT ToolKit

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.

Hope it's clear.
_________________
Intern  Joined: 30 Oct 2011
Posts: 30
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

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.

Check similar question: if-x-is-an-integer-does-x-have-a-factor-n-such-that-100670.html

Hope it's clear.

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)
Math Expert V
Joined: 02 Sep 2009
Posts: 64318
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

1
1
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.

Check similar question: if-x-is-an-integer-does-x-have-a-factor-n-such-that-100670.html

Hope it's clear.

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.
_________________
Intern  Joined: 20 Apr 2013
Posts: 7
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

2
crazypriya wrote:
Does the integer k have a factor p such that 1<p<k?
1.k>4!
2.13!+2<=k<=13!+13

OA B

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!
Math Expert V
Joined: 02 Sep 2009
Posts: 64318
Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

GMAT Club Legend  V
Joined: 11 Sep 2015
Posts: 4889
GMAT 1: 770 Q49 V46
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

3
1
gmatcracker2010 wrote:
i dont have any idea to solve the attached problem. Please provide methodology to solve such problems.

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: http://www.gmatprepnow.com/module/gmat-data-sufficiency?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

Cheers,
Brent
_________________
Director  P
Status: Professional GMAT Tutor
Affiliations: AB, cum laude, Harvard University (Class of '02)
Joined: 10 Jul 2015
Posts: 782
Location: United States (CA)
Age: 40
GMAT 1: 770 Q47 V48
GMAT 2: 730 Q44 V47 GMAT 3: 750 Q50 V42 GMAT 4: 730 Q48 V42
GRE 1: Q168 V169 WE: Education (Education)
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

Here is a visual that should help.
Attachments Screen Shot 2016-03-28 at 6.20.15 PM.png [ 133.71 KiB | Viewed 18461 times ]

Intern  Joined: 10 Dec 2014
Posts: 2
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

"Question basically asks whether k is a prime number. If it is, then it won't have a factor psuch that 1<p<k

(definition of a prime number)."

Hi,

I just don't understand this. Can you please explain?
Math Expert V
Joined: 02 Sep 2009
Posts: 64318
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

ruchitd wrote:
"Question basically asks whether k is a prime number. If it is, then it won't have a factor psuch that 1<p<k

(definition of a prime number)."

Hi,

I just don't understand this. Can you please explain?

A prime number is a positive integer with only two factors 1 and itself. So, a prime number does not have a factor which is more that 1 and less than itself.
_________________
Manager  B
Joined: 11 Feb 2017
Posts: 179
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

[quote="Bunuel"][quote="mneeti"][quote="Bunuel"]

Bunuel , Please make me understand one thing.

The question tells us that 1<p<k

which means p can be anything smaller than k , and question asks if p is a factor of K?

it means if k is 25 then is p a factor of 25? or we can assume p be any number less than 25 and check if its factor of 25 or not

My question is in option B,

If 13! + 2 <= k <= 13! + 13 , it means k is a very large digit and lies between those equalities.
SO NOW , p must be any integer less than k and we need to see if p is a factor of k?

How about 17? lets take p as 17

1<=17<=(13!+2,13!+3,13!+4........13!+13)
as we know from B that k has a factor , yes we can see 2, 3 ,4 ,5 till 13 it has factor but hat about 17?

17 satisfies the criteria and it should also be a factor and if its not then it must be uncertain if p is factor of k or not?

Math Expert V
Joined: 02 Sep 2009
Posts: 64318
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

rocko911 wrote:
Bunuel , Please make me understand one thing.

The question tells us that 1<p<k

which means p can be anything smaller than k , and question asks if p is a factor of K?

it means if k is 25 then is p a factor of 25? or we can assume p be any number less than 25 and check if its factor of 25 or not

My question is in option B,

If 13! + 2 <= k <= 13! + 13 , it means k is a very large digit and lies between those equalities.
SO NOW , p must be any integer less than k and we need to see if p is a factor of k?

How about 17? lets take p as 17

1<=17<=(13!+2,13!+3,13!+4........13!+13)
as we know from B that k has a factor , yes we can see 2, 3 ,4 ,5 till 13 it has factor but hat about 17?

17 satisfies the criteria and it should also be a factor and if its not then it must be uncertain if p is factor of k or not?

Forget about p. The question asks does a positive integer k, has a factor which is greater than 1 and less than k itself. As explained on previous pages this is the same as asking is k a prime number. If k is a prime number it won't have any such factor but if k is not a prime number, then it'll have such factor(s).
_________________
Intern  B
Joined: 18 May 2017
Posts: 47
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

1
Bunuel - i have subtracted 13! from each side of the inequality and got that k is equal or between 0 to 11. From this inequality i can't got the answer. Where am i wrong? By the way - i looked at the math book for inequality theory but haven't find something relevant. Is there any post that covers this topic?

Intern  B
Joined: 10 Jul 2017
Posts: 6
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

### Show Tags

Hi Bunuel,

I have the same question as oryahalom.

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

Thanks,
RD Re: Does the integer k have a factor p such that 1 < p < k ?   [#permalink] 12 Nov 2017, 08:22

Go to page    1   2    Next  [ 27 posts ]

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