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 350,000 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:

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 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! _________________

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.

Answer: B.

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....

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....

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....

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 ?

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....

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.

Re: If x is an integer, does x have a factor n such that 1 < [#permalink]
02 Oct 2013, 10:02

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. _________________

Re: If x is an integer, does x have a factor n such that 1 < [#permalink]
10 May 2015, 21:43

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. _________________

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

McCombs Acceptance Rate Analysis McCombs School of Business is a top MBA program and part of University of Texas Austin. The full-time program is small; the class of 2017...