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 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 2308:00 PM PDT
-09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
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.
26 Dec 2025, 22:59
Kudos
Bookmarks
a, b, c are consecutive but we do not know whether they are 11, 12, 13 or some other sequence
We are given a = b - 1
c = b + 1
a^c / b can be written as (b - 1)^(b + 1) / b
we know that, (b-1)^1 = -1 as remainder
(b-1)^2 = b^2 -2b + 1 = 1 as remainder
(b-1)^3 = b^3 - 3b(b-1) - 1 = -1 as remainder
The pattern for powers 1, 2, 3 is -1, 1, -1 which is -1 for odd and 1 for even, and this will continue
Also, since (b-1) = -1 (from above) the remainder can be 1 OR (b - 1)
Hence, Answer Option D
We are given a = b - 1
c = b + 1
a^c / b can be written as (b - 1)^(b + 1) / b
we know that, (b-1)^1 = -1 as remainder
(b-1)^2 = b^2 -2b + 1 = 1 as remainder
(b-1)^3 = b^3 - 3b(b-1) - 1 = -1 as remainder
The pattern for powers 1, 2, 3 is -1, 1, -1 which is -1 for odd and 1 for even, and this will continue
Also, since (b-1) = -1 (from above) the remainder can be 1 OR (b - 1)
Hence, Answer Option D
26 Dec 2025, 23:03
Kudos
Bookmarks
In this question we have given a. b. c being consecutive integers so we can write
a+2 = B+1 = c
Now we write equation a^c = b.Q + R
Lets open this equation
a can be written as b -1
(b - 1)^c = bQ+R
R should be between 0 and b
now if we see, or open left side
(-1)^c = then if
c is even then 1 would be the remainder and rest will be multiple of b
and if C is odd then -1 means b have to be there remainder can't be negative hence
b-1 will be it.
Hence I and II both is possible
Hence answer D
a+2 = B+1 = c
Now we write equation a^c = b.Q + R
Lets open this equation
a can be written as b -1
(b - 1)^c = bQ+R
R should be between 0 and b
now if we see, or open left side
(-1)^c = then if
c is even then 1 would be the remainder and rest will be multiple of b
and if C is odd then -1 means b have to be there remainder can't be negative hence
b-1 will be it.
Hence I and II both is possible
Hence answer D
26 Dec 2025, 23:36
Kudos
Bookmarks
Bunuel
We can definitely remove III as a choice, as the remainder when any number \(x\) is divided by \(b\) ranges between \(0\) and \(b-1\)
Now, since \(a < b < c\) and the three are consecutive integers, we can say:
\(\frac{a^1}{b}\) will always give a remainder \(a\), or \(b-1\), as we can then write \(a = 0*b + a\)
And, \(\frac{a^2}{b}\) will always give a remainder as 1, as \(a^2 = a*(b - 1) = ab - a = (a - 1 + 1)*b - a = (a - 1)*b + b - (b - 1) = (a - 1)*b + 1\)
Similarly, \(a^3 = a^2 * a = {(a - 1)*b + 1} * a = a*(a - 1)*b + a = a' * b + a\) (Assuming \(a' = a*(a - 1)\))
And \(a^4 = a^3 * a = a' * a * b + a^2 = a' * a * b + (a - 1)*b + 1 = (a' * a + (a - 1)) * b + 1 = a'' * b + 1\) (Assuming \(a'' = a' * a + (a - 1)\))
The pattern that we see is,
1. For all odd powers of a, \(\frac{a^c}{b}\) will give the remainder \(a\), or \(b-1\)
2. For all even powers of a, the remainder is 1
Therefore, in cases, when a,c are odd, the remainder is \(b-1\), while when a,c are even, the remainder is 1.
Hence, the answer is D
