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 2601:30 AM EDT
-02:30 AM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more.
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.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
05 Apr 2021, 00:28
Kudos
Bookmarks
Rule: when you divide a number by (10)^N power, the Remainder you get will be the Last N digits of that numbers
Therefore, if we divide (6)^17 by 100 ———> the remainder will give us the last 2 digits of the Dividend
(6)^4 = 1,296
We can Use the concept of negative remainders to make the calculation easier.
1300 is a multiple of 100.
Thus, the negative remainder when 1,296 is divided by 100 is ——-> -4
(6)^17 ——> (6^4)^4 * 6 ———> (1,296)^4 * 6
Divide each factor in the numerator by the Dividend of 100, multiplying the remainders to get the excess remainder
(1,296)^4 /100 ——-> Rem of = (-4)^4
*
6 /100 ——-> Rem of = 6
Excess remainder = (-4)^4 * 6 = 256 * 6
Dividing the excess remainder by 100:
256 /100 ——> Rem of = 56
*
6 /100 ———> Rem of = 6
Excess remainder = 56 * 6 = 336
336 /100 ———> yields a remainder of 36
The last 2 digits are 36
3 is the answer
Posted from my mobile device
Therefore, if we divide (6)^17 by 100 ———> the remainder will give us the last 2 digits of the Dividend
(6)^4 = 1,296
We can Use the concept of negative remainders to make the calculation easier.
1300 is a multiple of 100.
Thus, the negative remainder when 1,296 is divided by 100 is ——-> -4
(6)^17 ——> (6^4)^4 * 6 ———> (1,296)^4 * 6
Divide each factor in the numerator by the Dividend of 100, multiplying the remainders to get the excess remainder
(1,296)^4 /100 ——-> Rem of = (-4)^4
*
6 /100 ——-> Rem of = 6
Excess remainder = (-4)^4 * 6 = 256 * 6
Dividing the excess remainder by 100:
256 /100 ——> Rem of = 56
*
6 /100 ———> Rem of = 6
Excess remainder = 56 * 6 = 336
336 /100 ———> yields a remainder of 36
The last 2 digits are 36
3 is the answer
Posted from my mobile device
13 Sep 2024, 21:34
Kudos
Bookmarks
We need to find the tens digit of 6^17
Last Two digits of 6 follow following pattern
=> So, from 6^2 cycle of tens' digit repeats as 3, 1, 9, 7 , 5, 3, 1 9, 7 , 5
=> We have a cycle of 5 from the 2nd term
=> Tens' digit of 6^17 = Now 17 = 1(6^1 not part of cycle) + 15 (cycle of 5*3) + 1 (as cycle starts from 2nd term) = First term of the cycle = 3
So, Answer will be B
Hope it helps!
Link to Theory for Last Two digits of exponents here.
Link to Theory for Units' digit of exponents here.
Last Two digits of 6 follow following pattern
- Last Two Digits of 6^1 = 06
- Last Two Digits of 6^2 = 36
- Last Two Digits of 6^3 = 36*6 = 16
- Last Two Digits of 6^4 = 16*6 = 96
- Last Two Digits of 6^5 = 96*6 = 76
- Last Two Digits of 6^6 = 76*6 = 56
- Last Two Digits of 6^7 = 56*6 = 36 [ same as 6^2 ]
- Last Two Digits of 6^8 = 36*6 = 16 [ same as 6^3 ]
=> So, from 6^2 cycle of tens' digit repeats as 3, 1, 9, 7 , 5, 3, 1 9, 7 , 5
=> We have a cycle of 5 from the 2nd term
=> Tens' digit of 6^17 = Now 17 = 1(6^1 not part of cycle) + 15 (cycle of 5*3) + 1 (as cycle starts from 2nd term) = First term of the cycle = 3
So, Answer will be B
Hope it helps!
Link to Theory for Last Two digits of exponents here.
Link to Theory for Units' digit of exponents here.
