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.
Mar 2709:30 AM PDT
-10:30 AM PDT
Preparing for the GMAT? Avoid these 5 major mistakes that can lower your score and hurt your MBA dreams. In this expert panel discussion, 4 top GMAT coaches - GMAT Ninja, Jeff Miller from TTP, Rida Shafiq from eGMAT, and Chris Gentry from Manhattan Prep- Mar 27
10:00 AM PDT
-11:00 AM PDT
What happens after getting an MBA? Is it worth the investment? To find out, I interviewed 10 GMAT Club professionals who have completed their MBAs and built careers in consulting, finance, tech, and entrepreneurship. Their answers might surprise you! 
Mar 2911:30 AM EDT
-12:30 PM 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.
Mar 3005:30 AM PDT
-07:30 AM PDT
We know Strengthen and Weaken questions account for more than 50% of the CR questions on the GMAT. With CR becoming even more important on GMAT Focus, it's time you strengthen your weaknesses with an approach that improves your solving time and accuracy!
Mar 3011:00 AM IST
-01:00 PM IST
Attend this session to evaluate your current skill level, learn process skills, and solve through tough Arithmetic questions.
Mar 3110:00 AM EDT
-11:59 PM EDT
The Target Test Prep team has recently introduced TTP OnDemand GMAT, a 400-hour live masterclass video course created by Scott Woodbury-Stewart, founder and CEO of Target Test Prep.
Apr 0308:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
17 Sep 2024, 05:30
Kudos
Bookmarks
We need to find the remainder when 333^222 is divided by 7?
We solve these problems by using Binomial Theorem, where we split the number into two parts, one part is a multiple of the divisor(7) and a big number, other part is a small number.
=> \(333^{222}\) = \((329 + 4)^{222}\)
Watch this video to MASTER BINOMIAL Theorem
Now, if we use Binomial Theorem to expand this then all the terms except the last term will be a multiple of 7
=> All terms except the last term will give remainder of 0 when divided by 7
=> Problem is reduced to what is the remainder when the last term (i.e. 222C222 * 329^0 * 4^222) is divided by 7
=> Remainder of 4^222 is divided by 7
=> Remainder of 2^444 divided by 7
To solve this problem we need to find the cycle of remainder of power of 2 when divided by 7
Remainder of \(2^1\) (=2) by 7 = 2
Remainder of \(2^2\) (=4) by 7 = 4
Remainder of \(2^3\) (=8) by 7 = 1
Remainder of \(2^4\) (=16) by 7 = 2
Remainder of \(2^5\) (=32) by 7 = 4
Remainder of \(2^6\) (=64) by 7 = 1
=> Cycle is 3
Remainder of 444 by 3 = Remainder of (4+4+4) by 3 = 0
=> Remainder of \(2^{444}\) by 7 = Remainder of \(2^{Cycle}\) by 7 = Remainder of \(2^{3}\) by 7 = 1
So, Answer will be E
Hope it Helps!
Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem
We solve these problems by using Binomial Theorem, where we split the number into two parts, one part is a multiple of the divisor(7) and a big number, other part is a small number.
=> \(333^{222}\) = \((329 + 4)^{222}\)
Watch this video to MASTER BINOMIAL Theorem
Now, if we use Binomial Theorem to expand this then all the terms except the last term will be a multiple of 7
=> All terms except the last term will give remainder of 0 when divided by 7
=> Problem is reduced to what is the remainder when the last term (i.e. 222C222 * 329^0 * 4^222) is divided by 7
=> Remainder of 4^222 is divided by 7
=> Remainder of 2^444 divided by 7
To solve this problem we need to find the cycle of remainder of power of 2 when divided by 7
Remainder of \(2^1\) (=2) by 7 = 2
Remainder of \(2^2\) (=4) by 7 = 4
Remainder of \(2^3\) (=8) by 7 = 1
Remainder of \(2^4\) (=16) by 7 = 2
Remainder of \(2^5\) (=32) by 7 = 4
Remainder of \(2^6\) (=64) by 7 = 1
=> Cycle is 3
Remainder of 444 by 3 = Remainder of (4+4+4) by 3 = 0
=> Remainder of \(2^{444}\) by 7 = Remainder of \(2^{Cycle}\) by 7 = Remainder of \(2^{3}\) by 7 = 1
So, Answer will be E
Hope it Helps!
Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem
28 Dec 2024, 23:49
Kudos
Bookmarks
Which other numbers are like 7? For which numbers should we use Cyclic reminder approach and which numbers we should we Binomial approach. Please explain.
The units digit does not determine the remainder when dividing by 7. For example, 9 divided by 7 gives the remainder of 2, 19 divided by 7 gives the remainder of 5, 29 divided by 7 gives the remainder of 1, ...
Bunuel
gmatcracker2407
Hi,
I am confused between 2 approaches for these kinds of problems
Approach 1: Binomial Theorem.
Approach 2: Find the unit's digit of the exponent and then find the remainder.
Unit's digit of 333^222 = unit's digit of 3^222. Then divide that by 7.
Cyclicity of 3 = 4 {3,9,7,1}. 222/3 has a remainder of 2. 3^2 has a unit's digit of 9. 9/7 has a remainder of 2
In Approach 2, i don't always get the same ans as by using Approach 1. Which approach is preferred for these kinds of problems?
Thanks.
I am confused between 2 approaches for these kinds of problems
Approach 1: Binomial Theorem.
Approach 2: Find the unit's digit of the exponent and then find the remainder.
Unit's digit of 333^222 = unit's digit of 3^222. Then divide that by 7.
Cyclicity of 3 = 4 {3,9,7,1}. 222/3 has a remainder of 2. 3^2 has a unit's digit of 9. 9/7 has a remainder of 2
In Approach 2, i don't always get the same ans as by using Approach 1. Which approach is preferred for these kinds of problems?
Thanks.
The units digit does not determine the remainder when dividing by 7. For example, 9 divided by 7 gives the remainder of 2, 19 divided by 7 gives the remainder of 5, 29 divided by 7 gives the remainder of 1, ...
29 Dec 2024, 00:19
Kudos
Bookmarks
Jyoti1812
Which other numbers are like 7? For which numbers should we use Cyclic reminder approach and which numbers we should we Binomial approach. Please explain.
The units digit does not determine the remainder when dividing by 7. For example, 9 divided by 7 gives the remainder of 2, 19 divided by 7 gives the remainder of 5, 29 divided by 7 gives the remainder of 1, ...
Bunuel
gmatcracker2407
Hi,
I am confused between 2 approaches for these kinds of problems
Approach 1: Binomial Theorem.
Approach 2: Find the unit's digit of the exponent and then find the remainder.
Unit's digit of 333^222 = unit's digit of 3^222. Then divide that by 7.
Cyclicity of 3 = 4 {3,9,7,1}. 222/3 has a remainder of 2. 3^2 has a unit's digit of 9. 9/7 has a remainder of 2
In Approach 2, i don't always get the same ans as by using Approach 1. Which approach is preferred for these kinds of problems?
Thanks.
I am confused between 2 approaches for these kinds of problems
Approach 1: Binomial Theorem.
Approach 2: Find the unit's digit of the exponent and then find the remainder.
Unit's digit of 333^222 = unit's digit of 3^222. Then divide that by 7.
Cyclicity of 3 = 4 {3,9,7,1}. 222/3 has a remainder of 2. 3^2 has a unit's digit of 9. 9/7 has a remainder of 2
In Approach 2, i don't always get the same ans as by using Approach 1. Which approach is preferred for these kinds of problems?
Thanks.
The units digit does not determine the remainder when dividing by 7. For example, 9 divided by 7 gives the remainder of 2, 19 divided by 7 gives the remainder of 5, 29 divided by 7 gives the remainder of 1, ...
Check LAST DIGIT OF A POWER chapter here: https://gmatclub.com/forum/math-number- ... 88376.html
