GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Sep 2018, 20:29

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

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: 188
If x is an integer, does x have a factor n such that 1 <  [#permalink]

Show Tags

08 Sep 2010, 11:52
7
34
00:00

Difficulty:

15% (low)

Question Stats:

79% (00:38) correct 21% (01:17) wrong based on 865 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............
Math Expert
Joined: 02 Sep 2009
Posts: 49208
If x is an integer, does x have a factor n such that 1 <  [#permalink]

Show Tags

08 Sep 2010, 12:01
16
15
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) $$15!+2\leq{x}\leq{15!+15}$$ --> $$x$$ cannot 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.

_________________
Veritas Prep and Orion Instructor
Joined: 26 Jul 2010
Posts: 264

Show Tags

08 Sep 2010, 17:54
20
9
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.

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Veritas Prep Reviews

General Discussion
Manager
Joined: 17 Nov 2009
Posts: 223

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
Manager
Joined: 20 Jul 2010
Posts: 224

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: 188

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: 49208
If x is an integer, does x have a factor n such that 1 <  [#permalink]

Show Tags

09 Sep 2010, 14:37
1
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 cannot 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: 188

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: 49208

Show Tags

09 Sep 2010, 16:19
1
1
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.
_________________
Manager
Joined: 20 Oct 2013
Posts: 54
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

Intern
Joined: 06 Nov 2016
Posts: 30
Re: If x is an integer, does x have a factor n such that 1 <  [#permalink]

Show Tags

21 Feb 2017, 06:33
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.

Instead of multiplying all the numbers from 1 to 15, why not just take prime numbers between 1 and 15. That should still work, right?
Math Expert
Joined: 02 Sep 2009
Posts: 49208
Re: If x is an integer, does x have a factor n such that 1 <  [#permalink]

Show Tags

21 Feb 2017, 10:55
SVP482 wrote:
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.

Instead of multiplying all the numbers from 1 to 15, why not just take prime numbers between 1 and 15. That should still work, right?

No, it won't. Why should it? There are primes as well as non-primes from 1 to 15.
_________________
Intern
Joined: 28 Dec 2010
Posts: 23
If x is an integer, does x have a factor n such that 1 <  [#permalink]

Show Tags

29 Jun 2017, 15:27
1
Given: x is an integer
Question: Does x has a factor n such that 1<n<x? --> This a 'yes' or 'no' question

If x is not a prime number, then n (factor of x) should always be less equal than x.
Example: x = 8, factors of x i.e. n can be 1,2,4,8.

If x is a prime number, then n (factor of x) should always be 1, x
Example: x = 3, factors of x i.e n can be 1, 3

Statement 1: x > 4! => 4*3*2*1 = 24 now factors of x can be 1, 2, 3, 4, 6, 8, 12, 24. For the values of 1,2,3,4,6,12 it satisfies the condition we get 'yes' for the question but for 24 we get 'no'. Therefore, it is (not sufficient).

Statement 2: 15! + 2 ≤ x ≤ 15! + 15 --> range of value for x

15! = 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1
15! + 2 = 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1 + 2
= 2(15*14*13*12*11*10*9*8*7*6*5*4*3*1 + 1) ---> is completely divisible by 2 (2 is a the factor of 15!+2)

15! +15 =15(14*13*12*11*10*9*8*7*6*5*4*3*2*1 + 1) ---> is completely divisible by 15 (15 is a the factor15!+15)

From here we can see that for any value of x in the range it's factor will only be less than x.

Therefore, we can definately say "yes" that n will be less than x. (Sufficient)
Intern
Joined: 28 Feb 2015
Posts: 1
Re: If x is an integer, does x have a factor n such that 1 <  [#permalink]

Show Tags

01 Jul 2018, 21:56
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) $$15!+2\leq{x}\leq{15!+15}$$ --> $$x$$ cannot 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.

Hey Bunuel,

regarding statement 2: Let's say we would add a prime number bigger than 15 to the 15!, e.g. 17. Because it is not possible to factor out 17, the statement is not sufficient, right? Because the result might be prime or not. On the other hand, if we would add 16, we could still factor out 2 and 8 from 15! correct?

Best regards
chalmers15
Math Expert
Joined: 02 Sep 2009
Posts: 49208
Re: If x is an integer, does x have a factor n such that 1 <  [#permalink]

Show Tags

01 Jul 2018, 23:58
chalmers15 wrote:
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) $$15!+2\leq{x}\leq{15!+15}$$ --> $$x$$ cannot 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.

Hey Bunuel,

regarding statement 2: Let's say we would add a prime number bigger than 15 to the 15!, e.g. 17. Because it is not possible to factor out 17, the statement is not sufficient, right? Because the result might be prime or not. On the other hand, if we would add 16, we could still factor out 2 and 8 from 15! correct?

Best regards
chalmers15

You are almost right.

Yes, 15! + 16 is not a prime because we can factor out an integer greater than 1 from 15! + 16.
Yes, 15! + 17 might be a prime as well as not. But 15! + 17 is some specific number, so theoretically we could find out whether it's prime or not. If it were a prime, then $$15!+2\leq{x}\leq{15!+17}$$ would not be sufficient and if it were NOT a prime, then $$15!+2\leq{x}\leq{15!+17}$$ would be sufficient. Turns out that 15! + 17 is NOT a prime number: 15! + 17 = 1307674368017 = 9157×142805981. So, if (2) were $$15!+2\leq{x}\leq{15!+17}$$ it would still be sufficient. But you won't get such number on the GMAT because you'll need a computer to check whether such a large number is a prime (there are some particular values of $$x=n!+1$$ 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.).
_________________
Re: If x is an integer, does x have a factor n such that 1 < &nbs [#permalink] 01 Jul 2018, 23:58
Display posts from previous: Sort by

Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.