Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 26 Aug 2016, 07:36

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If x is an integer, does x have a factor n such that 1 <

Author Message
TAGS:

### Hide Tags

Manager
Joined: 16 Feb 2010
Posts: 225
Followers: 2

Kudos [?]: 242 [5] , given: 16

If x is an integer, does x have a factor n such that 1 < [#permalink]

### Show Tags

08 Sep 2010, 11:52
5
KUDOS
11
This post was
BOOKMARKED
00:00

Difficulty:

5% (low)

Question Stats:

76% (01:35) correct 24% (01:18) wrong based on 558 sessions

### HideShow timer Statistics

If x is an integer, does x have a factor n such that 1 < n < x?

(1) x > 3!

(2) 15! + 2 ≤ x ≤ 15! + 15

I am sure i ve seen a quesiton like this one before, but tot forgot how to solve it............
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 34449
Followers: 6267

Kudos [?]: 79544 [5] , given: 10022

### Show Tags

08 Sep 2010, 12:01
5
KUDOS
Expert's post
7
This post was
BOOKMARKED
zisis wrote:
If x is an integer, does x have a factor n such that 1 < n < x?

(1) x > 3!

(2) 15! + 2 ≤ x ≤ 15! + 15

I am sure i ve seen a quesiton like this one before, but tot forgot how to solve it............

If x is an integer, does x have a factor n such that 1 < n < x?

Question basically asks: is $$x$$ a prime number? If it is, then it won't have a factor $$n$$ such that $$1<n<x$$ (definition of a prime number).

(1) $$x>3!$$ --> $$x$$ is more than some number (3!). $$x$$ may or may not be a prime. Not sufficient.

(2) $$15!+2\leq{x}\leq{15!+15}$$ --> $$x$$ can not be a prime. For instance if $$x=15!+8=8*(2*3*4*5*6*7*9*10*11*12*13*14*15+1)$$, then $$x$$ is a multiple of 8, so not a prime. Same for all other numbers in this range: $$x=15!+k$$, where $$2\leq{k}\leq{15}$$ will definitely be a multiple of $$k$$ (as weould be able to factor out $$k$$ out of $$15!+k$$). Sufficient.

_________________
Manager
Joined: 17 Nov 2009
Posts: 239
Followers: 1

Kudos [?]: 36 [0], given: 17

### Show Tags

08 Sep 2010, 13:17
zisis wrote:
If x is an integer, does x have a factor n such that 1 < n < x?

(1) x > 3!

(2) 15! + 2 ≤ x ≤ 15! + 15

I am sure i ve seen a quesiton like this one before, but tot forgot how to solve it............

Rephrase: Is there a factor of x that is greater than 1? So x should not be prime

1. if x> 3! then x>6 but 7 is a prime and greater than 6 and 9 is not prine but greater than 6. No solid answer

2. If x lies between 15!+2 and 15!+ 15 then there are no primes here..

B
Veritas Prep GMAT Instructor
Joined: 26 Jul 2010
Posts: 236
Followers: 208

Kudos [?]: 431 [10] , given: 28

### Show Tags

08 Sep 2010, 17:54
10
KUDOS
8
This post was
BOOKMARKED
I love this question, so I'll chime in even if only to bump it to the top so more people see it!

Hopefully most can see pretty quickly that x > 3! just means x > 6, and that isn't nearly enough to tell us whether it is prime.

Statement 2 is pretty neat, though: 15! + 2 ≤ x ≤ 15! + 15

Think about 15!. 15! is 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1, which means that:

15! is a multiple of 2
15! is a multiple of 3
15! is a multiple of 4
etc....
15! is a multiple of 15

Now think about multiples of 2. Every SECOND number is a multiple of 2: 2, 4, 6, 8, 10. You create multiples of 2 by adding 2 to a previous multiple of 2. If a number is even, adding 2 "keeps it" even.

The same holds for 3. Every THIRD number (3, 6, 9, 12, 15, 18) is a multiple of 3. If you have a multiple of 3 and add 3 to it, it's still a multiple of 3.

This will hold for all of these numbers - because 15! is a multiple of every number between 2 and 15, then adding any number in that range of 2-15 will ensure that the new number remains a multiple of that number, and the new number will not be prime.

Therefore, statement 2 is sufficient, and the correct answer is B.

Now...consider the number 15! + 1. It's too big a number to know offhand whether it's prime, but we do know that it is NOT a multiple of any numbers 2-15. In order to be a multiple of 2, we'd have to add a multiple of 2 to 15!, and 1 breaks us off of that every-second-number cycle. Same for 3 - we'd need to add a multiple of 3 in order to keep 15! on that every-third cycle, so 15! is not a multiple of 3. We can prove that 15! + 1 is not divisible by any numbers between 2 and 15. It's largest prime factor must then be 17 or greater.

Understanding that ideology and being able to determine divisibility of large numbers can be quite helpful on questions that might otherwise seem impossible. Thanks for posting this question!
_________________

Brian

Save \$100 on live Veritas Prep GMAT Courses and Admissions Consulting

Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Senior Manager
Joined: 20 Jul 2010
Posts: 269
Followers: 2

Kudos [?]: 66 [0], given: 9

### Show Tags

09 Sep 2010, 06:26
Nice rephrase. Somehow in option 1 I took my values as 3!, 4!, 5! and thought they all will have a factor as needed.
_________________

If you like my post, consider giving me some KUDOS !!!!! Like you I need them

Manager
Joined: 16 Feb 2010
Posts: 225
Followers: 2

Kudos [?]: 242 [0], given: 16

### Show Tags

09 Sep 2010, 14:28
Bunuel wrote:
zisis wrote:
If x is an integer, does x have a factor n such that 1 < n < x?

(1) x > 3!

(2) 15! + 2 ≤ x ≤ 15! + 15

I am sure i ve seen a quesiton like this one before, but tot forgot how to solve it............

If x is an integer, does x have a factor n such that 1 < n < x?

Question basically asks: is $$x$$ a prime number? If it is, then it won't have a factor $$n$$ such that $$1<n<x$$ (definition of a prime number).

(1) $$x>3!$$ --> $$x$$ is more than some number (3!). $$x$$ may or may not be a prime. Not sufficient.

(2) $$15!+2\leq{x}\leq{15!+15}$$ --> $$x$$ can not be a prime. For instance if $$x=15!+8=8*(2*3*4*5*6*7*9*10*11*12*13*14*15+1)$$, then $$x$$ is a multiple of 8, so not a prime. Same for all other numbers in this range: $$x=15!+k$$, where $$2\leq{k}\leq{15}$$ will definitely be a multiple of $$k$$ (as weould be able to factor out $$k$$ out of $$15!+k$$). Sufficient.

I am able to follow until the point that I have highlighted red.......
tried to work your point my self so decided to find if x for 5!+3<x<5!+8 works with your statement above....

so -
$$5! = 120$$
$$5!+3 / 3 = 123 / 3 =41$$correct
$$5!+4 / 4 = 124 / 4 =31$$ correct
$$5!+5 / 5 = 125 / 5 =25$$ correct
$$5!+6 / 6 = 126 / 6 =21$$correct
$$5!+7 / 7 = 127 / 7 =18.14$$INCORRECT !!!!

so...? what am i doing wrong???
Math Expert
Joined: 02 Sep 2009
Posts: 34449
Followers: 6267

Kudos [?]: 79544 [1] , given: 10022

### Show Tags

09 Sep 2010, 14:37
1
KUDOS
Expert's post
zisis wrote:
I am able to follow until the point that I have highlighted red.......
tried to work your point my self so decided to find if x for 5!+3<x<5!+8 works with your statement above....

so -
$$5! = 120$$
$$5!+3 / 3 = 123 / 3 =41$$correct
$$5!+4 / 4 = 124 / 4 =31$$ correct
$$5!+5 / 5 = 125 / 5 =25$$ correct
$$5!+6 / 6 = 126 / 6 =21$$correct
$$5!+7 / 7 = 127 / 7 =18.14$$INCORRECT !!!!

so...? what am i doing wrong???

You replaced 15! by 5!. Thus you can not factor out 7 out of 5!+8 and this is the whole point here.

If you want to check with smaller numbers try $$5!+2\leq{x}\leq{5!+5}$$
_________________
Manager
Joined: 16 Feb 2010
Posts: 225
Followers: 2

Kudos [?]: 242 [0], given: 16

### Show Tags

09 Sep 2010, 15:49
Bunuel wrote:
zisis wrote:
I am able to follow until the point that I have highlighted red.......
tried to work your point my self so decided to find if x for 5!+3<x<5!+8 works with your statement above....

so -
$$5! = 120$$
$$5!+3 / 3 = 123 / 3 =41$$correct
$$5!+4 / 4 = 124 / 4 =31$$ correct
$$5!+5 / 5 = 125 / 5 =25$$ correct
$$5!+6 / 6 = 126 / 6 =21$$correct
$$5!+7 / 7 = 127 / 7 =18.14$$INCORRECT !!!!

so...? what am i doing wrong???

You replaced 15! by 5!. Thus you can not factor out 7 out of 5!+8 and this is the whole point here.

did it on excel and seems like you are right.....

15! 15!+x (15!+x)/x
1,307,674,368,000.00 1,307,674,368,003.00 435,891,456,001.00
1,307,674,368,000.00 1,307,674,368,004.00 326,918,592,001.00
1,307,674,368,000.00 1,307,674,368,005.00 261,534,873,601.00
1,307,674,368,000.00 1,307,674,368,006.00 217,945,728,001.00
1,307,674,368,000.00 1,307,674,368,007.00 186,810,624,001.00
1,307,674,368,000.00 1,307,674,368,008.00 163,459,296,001.00

now have to go back and understand how it works.....using the example taht I used (which you mentioned I should use smaller numbers), how do you know when x is too big for a and b, when k!+a<x<k!+b
?
Math Expert
Joined: 02 Sep 2009
Posts: 34449
Followers: 6267

Kudos [?]: 79544 [0], given: 10022

### Show Tags

09 Sep 2010, 16:19
Expert's post
1
This post was
BOOKMARKED
zisis wrote:
Bunuel wrote:
zisis wrote:
I am able to follow until the point that I have highlighted red.......
tried to work your point my self so decided to find if x for 5!+3<x<5!+8 works with your statement above....

so -
$$5! = 120$$
$$5!+3 / 3 = 123 / 3 =41$$correct
$$5!+4 / 4 = 124 / 4 =31$$ correct
$$5!+5 / 5 = 125 / 5 =25$$ correct
$$5!+6 / 6 = 126 / 6 =21$$correct
$$5!+7 / 7 = 127 / 7 =18.14$$INCORRECT !!!!

so...? what am i doing wrong???

You replaced 15! by 5!. Thus you can not factor out 7 out of 5!+8 and this is the whole point here.

did it on excel and seems like you are right.....

15! 15!+x (15!+x)/x
1,307,674,368,000.00 1,307,674,368,003.00 435,891,456,001.00
1,307,674,368,000.00 1,307,674,368,004.00 326,918,592,001.00
1,307,674,368,000.00 1,307,674,368,005.00 261,534,873,601.00
1,307,674,368,000.00 1,307,674,368,006.00 217,945,728,001.00
1,307,674,368,000.00 1,307,674,368,007.00 186,810,624,001.00
1,307,674,368,000.00 1,307,674,368,008.00 163,459,296,001.00

now have to go back and understand how it works.....using the example taht I used (which you mentioned I should use smaller numbers), how do you know when x is too big for a and b, when k!+a<x<k!+b
?

It seems that you don't understand the explanation. No need to check in excel: no integer $$x$$ satisfying $$15!+2\leq{x}\leq{15!+15}$$ will be a prime number.

There are 14 numbers satisfying it:
If $$x=15!+2$$ then we can factor out 2, so $$x$$ would be multiple of 2, thus not a prime;
If $$x=15!+3$$ then we can factor out 3, so $$x$$ would be multiple of 3, thus not a prime;
...

If $$x=15!+15$$ then we can factor out 15, so $$x$$ would be multiple of 15, thus not a prime.

Also 15!+{any multiple of prime less than or equal to 13} also won't be a prime number as we can factor out this prime (15! has all primes less than or equal to 13).

Now, for $$x=15!+1$$ or $$x=15!+17$$ or $$x=15!+19$$ we can not say for sure whether they are primes or not. In fact they are such a huge numbers that without a computer it's very hard and time consuming to varify their primality.

Hope it's clear.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11068
Followers: 511

Kudos [?]: 134 [0], given: 0

Re: If x is an integer, does x have a factor n such that 1 < [#permalink]

### Show Tags

02 Oct 2013, 11:02
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Manager
Joined: 20 Oct 2013
Posts: 66
Followers: 0

Kudos [?]: 2 [0], given: 27

Re: If x is an integer, does x have a factor n such that 1 < [#permalink]

### Show Tags

04 May 2014, 05:45
good qs!! and explainaton bunnel i am stalking you on GMATclub.com
_________________

Hope to clear it this time!!
GMAT 1: 540
Preparing again

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11068
Followers: 511

Kudos [?]: 134 [0], given: 0

Re: If x is an integer, does x have a factor n such that 1 < [#permalink]

### Show Tags

10 May 2015, 22:43
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11068
Followers: 511

Kudos [?]: 134 [0], given: 0

Re: If x is an integer, does x have a factor n such that 1 < [#permalink]

### Show Tags

13 Jun 2016, 23:52
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: If x is an integer, does x have a factor n such that 1 <   [#permalink] 13 Jun 2016, 23:52
Similar topics Replies Last post
Similar
Topics:
10 Dose positive integer x have a factor f such that 1<f<x ? 15 29 Dec 2014, 05:11
11 Does P have a factor X where 1<X<P , and X and P are positive integers 6 19 Nov 2014, 05:24
3 How many factors does x have, if x is a positive integer ? 7 17 Nov 2014, 12:29
6 Does integer n have 2 factors x & y such that 1 < x < y < n? 5 13 Jan 2014, 04:57
17 How many factors does the integer X have? 14 04 Oct 2010, 06:16
Display posts from previous: Sort by