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.
Apr 2308:00 PM PDT
-09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
Apr 2301:30 AM EDT
-02:30 AM EDT
Struggling with GMAT Verbal as a non-native speaker? Harsh improved his score from 595 to 695 in just 45 days—and scored a 99 %ile in Verbal (V88)! Learn how smart strategy, clarity, and guided prep helped him gain 100 points.
Apr 2512:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
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
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
42% (02:41) correct
58%
(02:33)
wrong
based on 677
sessions
History
Date
Time
Result
Not Attempted Yet
Is x divisible by 30?
(1) x = k*(m^3 - m), where m and k are both integers > 9
(2) x = n^5 - n, where n is an integer > 9
(1) x = k*(m^3 - m), where m and k are both integers > 9
(2) x = n^5 - n, where n is an integer > 9
As OA is not given, I tried it as below and for me the answer should be B. Any help in correcting or confirming the answer will be very much appreciated.
Question Says
is x divisible by 30 i.e. does x have atleast 2,3 and 5 as its Prime factors?
Considering Statement 1
x= k [m(m^2-1)] ------------------ (After factorization)------------------(1)
x = k (m-1,m and m+1) --------------------------------------------------(2)
From 2 we can tell that m-1,m and m+1 are consecutive integers. But say m = 17 then (m-1) (m) (m+1) is not divisible by 30 and also we don't know the value of K. But when m = 10 then (m-1) (m) (m+1) is divisible by 30 and irrespective of the value of K it will always be divisible by 30. Therefore this statement alone is insufficient to answer this question.
Considering statement 2
x = \(n^5\)-n
By factorization it can be simplified to \(n^2\) (n+1) (n-1). No matter whether n will be ODD or EVEN this will always be divisible by 30 because n^2 (n+1) (n-1) will always have more prime factors than 30. Therefore sufficient and hence B is my answer.
Question Says
is x divisible by 30 i.e. does x have atleast 2,3 and 5 as its Prime factors?
Considering Statement 1
x= k [m(m^2-1)] ------------------ (After factorization)------------------(1)
x = k (m-1,m and m+1) --------------------------------------------------(2)
From 2 we can tell that m-1,m and m+1 are consecutive integers. But say m = 17 then (m-1) (m) (m+1) is not divisible by 30 and also we don't know the value of K. But when m = 10 then (m-1) (m) (m+1) is divisible by 30 and irrespective of the value of K it will always be divisible by 30. Therefore this statement alone is insufficient to answer this question.
Considering statement 2
x = \(n^5\)-n
By factorization it can be simplified to \(n^2\) (n+1) (n-1). No matter whether n will be ODD or EVEN this will always be divisible by 30 because n^2 (n+1) (n-1) will always have more prime factors than 30. Therefore sufficient and hence B is my answer.
31 Jan 2012, 17:23
Kudos
Bookmarks
Is x divisible by 30?
(1) x = k*(m^3 - m), where m and k are both integers > 9 --> x=k*(m-1)m(m+1). Well, if k=30 then the answer is YES, but if k=11, then (m-1)m(m+1) only guarantees divisibility by 6, so the answer might be NO (for example m=12). Not sufficient.
(2) x = n^5 - n, where n is an integer > 9 --> the last digit of any integer in positive integer power repeats in pattern of 4 (some numbers have the pattern of 2 or 1, but pattern of 4 holds for them too). So, n^5 has the same last digit as n^1 or simply n, therefore the last digit of n^5-n is ALWAYS zero, so its ALWAYS divisible by 10. Next, n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=(n-1)n(n+1)(n^2+1) --> (n-1)n(n+1) is divisible by 3 (as there are 3 consecutive integers) --> n^5-n is divisible by 10*3=30. Sufficient.
Answer: B.
P.S. You have factored n^5-n incorrectly and also your reasoning for n^2(n+1)(n-1) is not right: it's not always divisible by 30, try n=12 for example.
Hope it's clear.
(1) x = k*(m^3 - m), where m and k are both integers > 9 --> x=k*(m-1)m(m+1). Well, if k=30 then the answer is YES, but if k=11, then (m-1)m(m+1) only guarantees divisibility by 6, so the answer might be NO (for example m=12). Not sufficient.
(2) x = n^5 - n, where n is an integer > 9 --> the last digit of any integer in positive integer power repeats in pattern of 4 (some numbers have the pattern of 2 or 1, but pattern of 4 holds for them too). So, n^5 has the same last digit as n^1 or simply n, therefore the last digit of n^5-n is ALWAYS zero, so its ALWAYS divisible by 10. Next, n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=(n-1)n(n+1)(n^2+1) --> (n-1)n(n+1) is divisible by 3 (as there are 3 consecutive integers) --> n^5-n is divisible by 10*3=30. Sufficient.
Answer: B.
P.S. You have factored n^5-n incorrectly and also your reasoning for n^2(n+1)(n-1) is not right: it's not always divisible by 30, try n=12 for example.
Hope it's clear.
27 May 2013, 11:56
Kudos
Bookmarks
enigma123
Bunuel has already explained the second part. Just another approach :
From F.S 1, we know that \(x = k*(m-1)*m*(m+1)\). Now, product of any 3 consecutive integers is ALWAYS divisible by 3.Also, the product of any two consecutive integers is ALWAYS divisible by 2. Thus, x is divisibe by 2 and 3--> x is divisible by atleast 6. Now we have to find out whether x is divisible by 5 or not.For k=m=10, it is. For k=m=13, it isn't.Insufficient.
From F.S 2, we know that x =\(n^5-n\) = \(n(n^2-1)(n^2+1)\) = \((n-1)(n)(n+1)(n^2+1)\). Just as above, we know that x is atleast divisibly by 6.Again, we have to find out whether it is divisibly by 5 or not-->If there was some way in which I could represent \((n^5-n)\) as a product of 5 consecutive integers, then I would be succesfull in proving that \(n^5-n\) is divisibly by 5 also.
Now, assuming n as the median,to get a product of 5 consecutive integers, we would need (n-2) and (n+2)-->\((n^2-4)\)-->\((n-1)n(n+1)[(n^2-4)+(4+1)]\) = (n-2)(n-1)n(n+1)(n+2) + 5(n-1)n(n+1)--> This will be always be divisible by 30.
There is actually a very famous theorem which states that \(x^p-x\) is always divisible by p, provided p is prime and x an integer. However, it is not relevant w.r.t GMAT.
