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

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Remainder of a huge exponent when divided by 7 [#permalink]
31 Aug 2011, 11:47

Step1:47 and 7 have no factors in common.. therefore we can use the Euler's number of 7(euler's number of a prime number is nothing but 1 less than the value of the number) which is 6. Step2:Now divide 203 by 6.. remainder is 5.. now the sum is 47 power 5 divided by 7 Step 3: 47 divided by 7 remainder is 5.. now the sum is 5 power 5 divided by 7. Step4: Answer is 3

Re: Remainder of a huge exponent when divided by 7 [#permalink]
31 Aug 2011, 20:29

The way to solve remainder - exponent questions is to find cyclicity of remainder. 47 ^1 Remainder is 5 47 ^2 = 47*47 Remainder when 5*5 is divided by 7 = 4 47 ^3 =(47*47)*47 Remainder when 4*5 is divided by 7 = 6 47 ^4 =(47*47)*(47*47) Remainder when 4*4 is divided by 7 = 2 47 ^5= Remainder when 5*2 is divided by 7 = 3 47^6 = Remainder when 5*3 is divided by 7 = 1 47^7 = remainder when 1*5 is divided by 7 = 5

We can see that after 6 the remainders will follow the same cycle. Thus remaniders will repaeat themselves with a cyclicity of 6. For 203, there would be 33 full ccyles of 6 and 5 remaining, so the remainder will be same as 47^5 = 3

Re: Remainder of a huge exponent when divided by 7 [#permalink]
31 Aug 2011, 22:33

x to the power y when divided by z what is the remainder? Consider this as the sample question. Now if x and z have no factors in common other than 1, that is if x and z are co-primes then follow the steps given below: 1. Find the euler's number of z. Euler's number of a prime number say z is nothing but z-1. 2. Now divide 'y' by the euler's number 'z-1' , let the remainder be 'r'. 3. Divide 'x' by z and let the remainder be q. 4. Now the question boils down to 'q' to the power 'r' divided by z .

Re: Remainder of a huge exponent when divided by 7 [#permalink]
31 Aug 2011, 22:51

1

This post received KUDOS

kuttingchai.. in the example 47 and 7 are co-primes.. 7 is a prime number. So find euler's number of 7, which is 6. Now divide 203 by 6..remainder is 5. Now we can replace the power 203 with 5. The question is now 47 to the power 5 when divided by 7 i.e.. 47*47*47*47*47/7. 47/7 remiander is 5. So the question now becomes 5*5*5*5*5/7.. answer is 3.

Hope this helps u.

This is not applicable to the problem 546^100 divided by 7.. because 546 and 7 are not co-primes.

In general the euler's number of N = N*(1-(1/a))*(1-(1/b)).... where a,b.. are the prime factors of N. Example.. to find euler number of 18, 18 = 2*3*3(expressing in terms of prime factors) Euler's no = 18*(1-(1/2))(1-(1/3)) = 18*(1/2)*(2/3) = 6. euler's no of 18 is 6.

Re: Remainder of a huge exponent when divided by 7 [#permalink]
01 Sep 2011, 04:23

Lets see what we get by doing it by binomial theorm. the theorem states that (a + b)^n = nC0a^n + nC1a^(n − 1)b + nC2a^(n − 2)b^2 + ... + nCnb^n Lets assume here a= 49 and b= -2 and n =203 so 47^203 is essentially (49-2)^203. Now if we expand in other than the last term i.e nCnb^n every other term has a i.e. 49 at different powers. So these all are divisible by 7. So we have 2^203 which is 8^67*2^2 if we divide 8 by 7 the remainder is 1 so we have 2^2/7 so remainder is 4

Re: Remainder of a huge exponent when divided by 7 [#permalink]
01 Sep 2011, 05:46

I get a remainder of 0.

Every 47 will give remainder of 5. So 47 times 203 will give us remainder of 5 x 203 = 1015. Now we need to divide 1015 by 7 again to check out how many 7's are there in 1015. Since 1015 is divisible by 7, 47^203 is divisble by 7.

Re: Remainder of a huge exponent when divided by 7 [#permalink]
01 Sep 2011, 07:21

tssambi, Every 47 will give remainder of 5. So 47 times 203 will not give us remainder of 5 x 203 = 1015; it will give us 5^203. So this calculation is wrong

Re: Remainder of a huge exponent when divided by 7 [#permalink]
01 Sep 2011, 07:58

Sorry my previous answer is wrong It should not be 4 it should be 3. Here we are using subtraction in the binomial theorem. So this answer should be base on the following theory. lets consider 23/7. if we do it by addition i.e 23/7= (21+2)/7 (a=21,b=2) then applying the binomial it is remainder of 21/7+ remainder of 2/7= remainder of 23/7 21/7 is 0 so 2/7 or 2 is the reminder. Now lets do it with subtraction 23/7= (28-5)/7 (a=28,b=5) then applying binomial remainder of 28/7 - remainder of 5/7 = remainder of 23/7 28/7 is 0 -5/7 = -5.

so we should subtract it from the divisor which is 7. so the remainder is again 7-5 =2

So in my previous answer the theorem remains the correct; only key point is if we are using subtraction in the binomial; we must subtract the end result from the divisor . i.e. in this case 7-4 = 3

gmatclubot

Re: Remainder of a huge exponent when divided by 7
[#permalink]
01 Sep 2011, 07:58

Wow...I'm still reeling from my HBS admit . Thank you once again to everyone who has helped me through this process. Every year, USNews releases their rankings of...

Almost half of MBA is finally coming to an end. I still have the intensive Capstone remaining which started this week, but things have been ok so far...