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.
31 Jan 2012, 16:25
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
64% (02:12) correct
36%
(02:31)
wrong
based on 1972
sessions
History
Date
Time
Result
Not Attempted Yet
If x^3 – x = n and x is a positive integer greater than 1, is n divisible by 8?
(1) When 3x is divided by 2, there is a remainder.
(2) x = 4y + 1, where y is an integer.
(1) When 3x is divided by 2, there is a remainder.
(2) x = 4y + 1, where y is an integer.
I will really appreciate if you can tell me whether I am right or wrong:
I have gone for D as an answer.
Let's simply the question stem a bit:
\(x^3\) - x = n which will give x (\(x^2\)-1) = n which will give (x-1), x and (x+1) = n. Therefore n is a product of 3 consecutive integers.
Now moving on to the statements
Statement 1 implies that x is ODD. Because ODD/EVEN will always end with some remainder.
X is ODD which means that the product will always be divisible by 2 three times and therefore will be divisible by 8.
Statement 2 implies that when x is divided by 4 it leaves a remainder 1 i.e. x is ODD. which is same as above and therefore sufficient.
Therefore for me both statements alone are sufficient to answer this question.
I have gone for D as an answer.
Let's simply the question stem a bit:
\(x^3\) - x = n which will give x (\(x^2\)-1) = n which will give (x-1), x and (x+1) = n. Therefore n is a product of 3 consecutive integers.
Now moving on to the statements
Statement 1 implies that x is ODD. Because ODD/EVEN will always end with some remainder.
X is ODD which means that the product will always be divisible by 2 three times and therefore will be divisible by 8.
Statement 2 implies that when x is divided by 4 it leaves a remainder 1 i.e. x is ODD. which is same as above and therefore sufficient.
Therefore for me both statements alone are sufficient to answer this question.
31 Jan 2012, 16:34
Kudos
Bookmarks
enigma123
Yes, your reasoning is correct.
If x^3 - x = n and x is a positive integer greater than 1, is n divisible by 8?
x^3-x=x(x^2-1)=(x-1)x(x+1), notice that we have the product of three consecutive integers. Now, if x=odd, then (x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8 (basically if x=odd then (x-1)x(x+1) will be divisible by 8*3=24).
(1) When 3x is divided by 2, there is a remainder --> 3x=odd --> x=odd --> (x-1)x(x+1) is divisible by 8. Sufficient.
(2) x = 4y + 1, where y is an integer --> x=even+odd=odd --> (x-1)x(x+1) is divisible by 8. Sufficient.
Answer: D.
Hope it helps.
04 Dec 2013, 12:47
Kudos
Bookmarks
enigma123
Statement 1:
3x/2 gives remainder . This means x is odd.
odd^3 = odd and odd- odd = even.
x >1 and an odd integer . Lets take x = 3
n = x^3 - x = 27-3
n= 24. 24/8 ->Remainder = 0
Lets take x = 5
n = x^3 - x = 125-5
n= 120. 120/8 ->Remainder = 0
Sufficient
Statement 2:
x = 4y+1 . y->integer
Put x in equation n = x^3 - x
n = (4y+1)^3 - (4y+1)
apply formula(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
(4y+1)^3 = 64y^3 + 3*16x^2 + 3*4y*1 + 1 = 64y^3 + 48x^2 + 12y + 1
n = 64y^3 + 48x^2 + 12y + 1 - 4y - 1
n = 64y^3 + 48x^2 + 8y
n = 8(8y^3 + 6x^2+1)
Hence n is divisible by 8. -> Sufficient
Hence D
