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 500,000 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:

What is the remainder when (18^22)^10 is divided by 7 ?

А 1 B 2 C 3 D 4 E 5

I think this question is beyond the GMAT scope. It can be solved with Fermat's little theorem, which is not tested on GMAT. Or another way:

\((18^{22})^{10}=18^{220}=(14+4)^{220}\) now if we expand this all terms but the last one will have 14 as multiple and thus will be divisible by 7. The last term will be \(4^{220}\). So we should find the remainder when \(4^{220}\) is divided by 7.

\(4^{220}=2^{440}\).

2^1 divided by 7 yields remainder of 2; 2^2 divided by 7 yields remainder of 4; 2^3 divided by 7 yields remainder of 1;

2^4 divided by 7 yields remainder of 2; 2^5 divided by 7 yields remainder of 4; 2^6 divided by 7 yields remainder of 1; ...

So the remainder repeats the pattern of 3: 2-4-1. So the remainder of \(2^{440}\) divided by 7 would be the same as \(2^2\) divided by 7 (440=146*3+2). \(2^2\) divided by 7 yields remainder of 4.

When we expand \((7 + 1)^{146}\), we get lots of terms such that \(7^{146}\) is the first term and 1 is the last term. If you multiply \((7 + 1)^{146}\) by 4, you get the same terms except each is multiplied by 4 so the first term is \(4*7^{146}\) and the last one is 4. Every term will still have a 7 in it except the last term. Since the last term is 4, the remainder will be 4.
_________________

Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]

Show Tags

10 Jun 2013, 03:50

1

This post received KUDOS

Financier wrote:

What is the remainder when (18^22)^10 is divided by 7 ?

А. 1 B. 2 C. 3 D. 4 E. 5

^ 1.(18)^220 = (((18) ^4)^5)^11 2. When 18 is divided by 7 the remainder is 4 3. Now 4^4 is 256. when divided by 7, the remainder is 4 4. Since the remainder is again 4, compute 4^5 = 1024. when divided by 7, the remainder is 2 5. Since 2 is the remainder, now we compute 2^11 = 2048. When divided by 7, the remainder is 4. This is the answer.

Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]

Show Tags

17 May 2014, 07:23

1

This post received KUDOS

Bunuel wrote:

gaurav1418z wrote:

Bunuel

(18^22)^10 = 18^220

I remember you quoted that for (xyz)^n, if we are asked to find the remainder, we can find remainder for z^n

so i found remainder for 8^220, and got answer as 1

Where am i going wrong?

Where did I write that? I think that you mean the following part from Number Theory booksaying that the last digit of \((xyz)^n\) is the same as that of \(z^n\). But the last digit of a number does not determine its remainder upon dividing by 7. For example, 8 divided by 7 gives the remainder of 1 while 18 divided by 7 gives the remainder of 4.

Another approach to solve such kind of questions is using cyclicity. Lemme try and explain this.

We have (18^22)^10 = 18^220

Now we try to see what are the remainders for various powers of 18 when it is divided by 7

18^1 leaves 4 as remainder 18^2 leaves 2 as remainder 18^3 leaves 1 as remainder 18^4 leaves 4 as remainder

Hence, we see after 3 set of powers, the remainder starts repeating. Now, this means if the power is multiple of 3, the remainder is 1, if the power leaves remainder 1 when divided by 3, the actual answer will be 4, and so forth.

We see, when 220 is divided by 3, we get 1 as remainder, showcasing 18^1 case, hence the final remainder is 4.

This is because 18^220 = 18^219 x 18 = (18^3)^73 x 18

Thus, we need to find the remainder when 18 is divided by 4 as 18^219 gives 1 as remainder.

Just remember that all terms in the expression (a+b)^n are divisible by a except for the last term i.e b^n. My solution uses just this much.

I need someone to validate this approach or did I just go absolutely berserk

Yes, that's your binomial theorem concept applied to remainders. In case you are interested in checking out the details of this approach, look at this post: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/ _________________

Bunuel what is wrong with the approach below- 18^220 = 2^220 3^440 this product will have last digit 6... Can't we use cyclisity or some other approach to solve? On the face of it it looks like a GMAT type problem...

What is the remainder when (18^22)^10 is divided by 7 ?

А 1 B 2 C 3 D 4 E 5

I think this question is beyond the GMAT scope. It can be solved with Fermat's little theorem, which is not tested on GMAT. Or another way:

\((18^{22})^{10}=18^{220}=(14+4)^{220}\) now if we expand this all terms but the last one will have 14 as multiple and thus will be divisible by 7. The last term will be \(4^{220}\). So we should find the remainder when \(4^{220}\) is divided by 7.

\(4^{220}=2^{440}\).

2^1 divided by 7 yields remainder of 2; 2^2 divided by 7 yields remainder of 4; 2^3 divided by 7 yields remainder of 1;

2^4 divided by 7 yields remainder of 2; 2^5 divided by 7 yields remainder of 4; 2^6 divided by 7 yields remainder of 1; ...

So the remainder repeats the pattern of 3: 2-4-1. So the remainder of \(2^{440}\) divided by 7 would be the same as \(2^2\) divided by 7 (440=146*3+2). \(2^2\) divided by 7 yields remainder of 4.

Answer: D.

Hi Karishma/Bunnel,

Can you point out the mistake here: R(2^440)/7:

2^440 = 2*2^339 = 2*8^113=2*(7+1)^113=2*(7*I+1^113), here I is an integer.

Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]

Show Tags

22 Apr 2014, 18:07

(18^{22})^{10}=18^{220}=(7+11)^{220}. If we expand this equation all terms will be divisible by 7 except the last one.

The last one will be 11^{220}. So we should find the remainder when 11^{220} is divided by 7.

11^{220}=?

11^1 divided by 7 yields remainder of 4; 11^2 divided by 7 yields remainder of 2; 11^3 divided by 7 yields remainder of 1; 11^4 divided by 7 yields remainder of 4; Now we have a pattern 4,2,1,4,2,1

Conclusion: the remainder repeats the pattern of 3: 4-2-1. So the remainder of 11^{220} divided by 7 would be the same as 11*1 (that is because 220 is 73*3+1) Answer: D.

I remember you quoted that for (xyz)^n, if we are asked to find the remainder, we can find remainder for z^n

so i found remainder for 8^220, and got answer as 1

Where am i going wrong?

Where did I write that? I think that you mean the following part from Number Theory book (math-number-theory-88376.html) saying that the last digit of \((xyz)^n\) is the same as that of \(z^n\). But the last digit of a number does not determine its remainder upon dividing by 7. For example, 8 divided by 7 gives the remainder of 1 while 18 divided by 7 gives the remainder of 4.
_________________

How did you get 12 as remainder when -1 is divided by 13, i didnt quite understand how did you substitute quotient and the remainder in the formula below

How did you get 12 as remainder when -1 is divided by 13, i didnt quite understand how did you substitute quotient and the remainder in the formula below

-1 = 13(-1) + 12

Hi there,

Remember that remainder can never be negative so when you get remainder negative as in above case then you add divisor to it

So we have remainder -1+13 =12

Posted from my mobile device _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...