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 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.
09 Jan 2023, 10:00
Kudos
Bookmarks
D
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:
47% (02:46) correct
53%
(02:34)
wrong
based on 138
sessions
History
Date
Time
Result
Not Attempted Yet
If n is a positive integer, which of the following must be divisible by 12?
I. \(n^4–n^2\)
II. \((n-6)^2(n-4)(n+13)^2\)
III. \(2n(n-10)(n+13)\)
A. I and II only
B. II and III only
C. I and III only
D. I, II and III
E. None of the above
I. \(n^4–n^2\)
II. \((n-6)^2(n-4)(n+13)^2\)
III. \(2n(n-10)(n+13)\)
A. I and II only
B. II and III only
C. I and III only
D. I, II and III
E. None of the above
09 Jan 2023, 14:08
Kudos
Bookmarks
Bunuel
To determine whether a number or term is divisible by 12, we must ensure that the number / term is divisible by 4 and is divisible by 3.
1. \(n^4 -n^2\)
\(n^2(n^2-1)\)
\(n^2(n+1)(n-1)\)
\(n*(n-1)n(n+1)\)
Check for divisibility by 4:
Case 1: n is even
As we have two even terms in the product, the number is divisible by 4.
Case 2: n is odd
When n is odd, (n-1) & (n+1) ⇒ even
As we have two even terms in the product, the number is divisible by 4.
Check for divisibility by 3: We have product of three consecutive terms, therefore the number will be divisible by 3.
Conclusion: As \(n^4 -n^2\) is divisible by both 4 and by 3, it is divisible by 12
2. \((n-6)^2 (n-4)(n+13)^2\)
We can use the knowledge of remainders and even and odd integers to do so.
We will use the concept of even / odd to check the divisibility of 4, and the concept of remainders to check the divisibility of 3.
Check for divisibility by 4:
Case 1 : n is even
even^2 * even * odd^2
As we have two even terms in the product, we can be sure that the number is divisible by 4.
Case 2 : n is odd
odd^2 * odd * even^2
Again, we see that the product has two even terms in the product, hence the number is divisible by 4
In both the above cases, the number is divisible by 4.
Check for divisibility by 3:
We can use the concept of remainders to check the divisibility of 3. "n" when divided by 3, can have a remainder 0, 1 or 2. Let's check if the term is divisible by 3 in each of the case.
Case 1: Remainder(n/3) = 0; i.e. n is a multiple of 3
If n is a multiple of 3, \((n-6)^2\) will be divisible by 3. Hence the entire term is divisible by 3.
Case 2: Remainder(n/3) = 1; i.e. n when divided by 3, leaves a remainder = 1
Remainder(\(\frac{(n-6)^2 (n-4)(n+13)^2 }{ 3} \)) = \((1-0)^2 (1-1)(1+1)^2\) = 0
In this case (n-4) will be divisible by 3, hence the entire term is divisible by 3.
Case 3: Remainder(n/3) = 2; i.e. n when divided by 3, leaves a remainder = 2
Remainder(\(\frac{(n-6)^2 (n-4)(n+13)^2 }{ 3} \)) = \((2-0)^2 (2-1)(2+1)^2\)
In this case, the last term \((n+13)^2\) will be divisible by 3.
In all the above cases, the number is divisible by 3.
Conclusion:As \( (n-6)^2 (n-4)(n+13)^2\) is divisible by both 4 and by 3, it is divisible by 12
3. \(2n (n-10) (n+13)\)
We can use the same approach in this as well. We can use the concept of even and odd to determine the divisibility by 4 and use the concept of remainders to determine the divisibility by 3
Check for divisibility by 4:
Case 1: n is even
If n is even, 2n is divisible by 4. Hence the entire term is divisible by 4.
Case 2: n is odd
If n is odd, (n+13) is even. We already have 2 in the equation, so the presence of an even term guarantees that the term will be divisible by 4.
Hence in both the cases, the number is divisible by 4.
Check for divisibility by 3:
As in previous case, we can use the concept of remainder to check the divisibility of 3.
Case 1: Remainder(n/3) = 0; i.e. n is a multiple of 3
2n (n-10) (n+13) is divisible by 3.
Case 2: Remainder(n/3) = 1; i.e. n when divided by 3, leaves a remainder = 1
Remainder (\(\frac{2n(n-10)(n+13)}{3}\)) = 2*1*(1-1)(1+1)
In this case the term (n-10) will be divisible by 3.
Case 3: Remainder(\(\frac{n}{3}\)) = 2; i.e. n when divided by 3, leaves a remainder = 2
Remainder (\(\frac{2n(n-10)(n+13)}{3}\)) = 2*2*(2-1)(2+1)
In this case the term (n+ 13 ) will be divisible by 3.
In all the above cases, the number is divisible by 3.
Conclusion: As \(2n (n-10) (n+13)\) is divisible by both 4 and by 3, it is divisible by 12
Option D
15 Jul 2025, 01:16
Kudos
Bookmarks
Hey, i dont really get why it MUST be divisible by 12 for any of the three equations.
If i make n=1, the first one is not divisible by 12,
similar if i make any of the brackets 0 in the second and third equation they would also be not divisible by 12.
all the n values would be positive integers, so why is the answer not E?
If i make n=1, the first one is not divisible by 12,
similar if i make any of the brackets 0 in the second and third equation they would also be not divisible by 12.
all the n values would be positive integers, so why is the answer not E?
