Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
08 Sep 2010, 11:52
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
70% (00:59) correct
30%
(01:43)
wrong
based on 1560
sessions
History
Date
Time
Result
Not Attempted Yet
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
(1) x > 3!
(2) 15! + 2 ≤ x ≤ 15! + 15
08 Sep 2010, 17:54
Kudos
Bookmarks
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!
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!
08 Sep 2010, 12:01
Kudos
Bookmarks
zisis
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.
Answer: B.
i am stalking you on GMATclub.com 
