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.
02 Apr 2020, 05:06
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
60% (02:03) correct
40%
(02:01)
wrong
based on 460
sessions
History
Date
Time
Result
Not Attempted Yet
Find the remainder when \(25^{18}\) is divided by 9.
A. 1
B. 2
C. 3
D. 5
E. 6
A. 1
B. 2
C. 3
D. 5
E. 6
02 Apr 2020, 05:13
Kudos
Bookmarks
Dillesh4096
\(R[\frac{25^{18}}{9}] = R[\frac{(27-2)^{18}}{9}] = R[\frac{(-2)^{18}}{9}] = R[\frac{(2)^{18}}{9}] \)
\(R[\frac{(2)^{3}}{9}] = 8 or -1\) (negative remainder)
CONCEPT: Remainder can be written in both +ve and -ve form for simplification. When divisor is 9 and remainder is 7 then the number has either 7 in excess or number is 2 short to be divisible by 9 hence
Remainder (+7) = Remainder (-2) when divisor is 9
Remainder (+7) = Remainder (-2) when divisor is 9
Taking exponent 6 both sides
\(R[\frac{(2)^{18}}{9}] = (-1)^6 = 1 \)
Answer: Option A
The concept video of remainder theorem is as follows:
27 Sep 2020, 08:57
Kudos
Bookmarks
A method which can be used in such a case is using cyclicity of remainders. In this we see how many cycles are required to get a remainder of 1.
Interestingly, if you get a remainder of -1, then the double of the cycles gives a remainder of 1.
We get a remainder of -1, if the remainder is 1 less than the divisor (eg, 31 divided by 8 gives a remainder of 7 which is also -1)
We stop the cycle, when we get a remainder of 1 or 8. If we get a remainder of 8, then double that cycle will give a remainder of 1.
\(R[\frac{(25)^{18}}{9}] = R[\frac{5^{36}}{9}]\)
Cycle 1: \(R[\frac{5}{9}] = 5\)
Cycle 2: We multiply the remainder of the previous cycle by 5 and get the next remainder. \(R[\frac{5 * 5}{9}] = R[\frac{25}{9}] = 7\)
Cycle 3: \(R[\frac{5 * 7}{9}] = R[\frac{35}{9}] = 8 \space or -1\)
Therefore, we shall get a remainder of 1 in the 6th cycle. This means that the remainder for every 6 powers of 5 divided by 9 = 1
\(R[\frac{5^6}{9}] = 1\)
Therefore \(R[\frac{5^{36}}{9}] = R[\frac{(5^6)^6}{9}] = R[\frac{1}{9}] = 1\)
Option A
Arun Kumar
Interestingly, if you get a remainder of -1, then the double of the cycles gives a remainder of 1.
We get a remainder of -1, if the remainder is 1 less than the divisor (eg, 31 divided by 8 gives a remainder of 7 which is also -1)
We stop the cycle, when we get a remainder of 1 or 8. If we get a remainder of 8, then double that cycle will give a remainder of 1.
\(R[\frac{(25)^{18}}{9}] = R[\frac{5^{36}}{9}]\)
Cycle 1: \(R[\frac{5}{9}] = 5\)
Cycle 2: We multiply the remainder of the previous cycle by 5 and get the next remainder. \(R[\frac{5 * 5}{9}] = R[\frac{25}{9}] = 7\)
Cycle 3: \(R[\frac{5 * 7}{9}] = R[\frac{35}{9}] = 8 \space or -1\)
Therefore, we shall get a remainder of 1 in the 6th cycle. This means that the remainder for every 6 powers of 5 divided by 9 = 1
\(R[\frac{5^6}{9}] = 1\)
Therefore \(R[\frac{5^{36}}{9}] = R[\frac{(5^6)^6}{9}] = R[\frac{1}{9}] = 1\)
Option A
Arun Kumar
