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:

Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]

Show Tags

31 Aug 2013, 00:13

Rem(32^32^32)=Rem(4^32)^32 now 4^3 = 64 = 63+1. hence when 64 to the power anything is divided by 7, the remainder will always be 1 so, Rem(32^32)/3 = Rem(33-1)^32/3 = 1. Hence Rem(4*64^k)/7 = 4

What is the remainder when \(32^{32^{32}}\) is divided by 7?

A. 5 B. 4 C. 2 D. 0 E. 1

Please do not just post the answer, do explain as well.

I will post the Answer and the explanation after some replies.

If we use the above approach I'd work with prime as a base.

\(32^{{32}^{32}}=(28+4)^{{32}^{32}}\) now if we expand this, all terms but the last one will have 28 as a multiple and thus will be divisible by 7. The last term will be \(4^{{32}^{32}}=4^{{(2^5)}^{32}}=4^{2^{160}}=2^{2^{161}}\). So we should find the remainder when \(2^{2^{161}}\) is divided by 7.

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; ...

The remainder repeats the pattern of 3: 2-4-1.

So we should find \(2^{161}\) (the power of 2) is 1st, 2nd or 3rd number in the above pattern of 3. \(2^{161}\) is 2 in odd power, 2 in odd power gives remainder of 2 when divided by cyclicity number 3, so it's the second number in pattern. Which means that remainder of \(2^{2^{161}}\) divided by 7 would be the same as \(2^2\) divided by 7. \(2^2\) divided by 7 yields remainder of 4.

Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]

Show Tags

04 Feb 2014, 22:40

Is there a post or resource about remainder cyclicities in general ? I know already about the units cyclicity but i am hoping for one about remainder cyclicities .

When we divide the power 5(32)(32) by three we get: + 2 remainder hence the remainder will be the second one in the series which is 4.
_________________

Unlock a set of 15 strategy videos that can help you plan your GMAT prep better! Sign up here to get complete access: https://goo.gl/OyUtLO

Do you find yourself struggling with questions on Permutations and Combinations? Attend this webinar to learn quick and non-formulae approach to counting methods! Register here: https://goo.gl/kJ7na7

For more info on GMAT and MBA, follow us on @AskCrackVerbal

What is the remainder when \(32^{32^{32}}\) is divided by 7?

A. 5 B. 4 C. 2 D. 0 E. 1

Please do not just post the answer, do explain as well.

I will post the Answer and the explanation after some replies.

If we use the above approach I'd work with prime as a base.

\(32^{{32}^{32}}=(28+4)^{{32}^{32}}\) now if we expand this, all terms but the last one will have 28 as a multiple and thus will be divisible by 7. The last term will be \(4^{{32}^{32}}=4^{{(2^5)}^{32}}=4^{2^{160}}=2^{2^{161}}\). So we should find the remainder when \(2^{2^{161}}\) is divided by 7.

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; ...

The remainder repeats the pattern of 3: 2-4-1.

So we should find \(2^{161}\) (the power of 2) is 1st, 2nd or 3rd number in the above pattern of 3. \(2^{161}\) is 2 in odd power, 2 in odd power gives remainder of 2 when divided by cyclicity number 3, so it's the second number in pattern. Which means that remainder of \(2^{2^{161}}\) divided by 7 would be the same as \(2^2\) divided by 7. \(2^2\) divided by 7 yields remainder of 4.

Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]

Show Tags

12 Jun 2014, 22:28

Brunel,

Can we solve it by expanding 32 as (35-3) ?

Here is how I approached : 32^32^32= (35-3) ^32^32

All other terms are divisible by 7 , barring last term 3^32^32. Here is the cyclicity of

3^1 when divided by 7 gives a remainder of 3 3^2 when divided by 7 gives a remainder of 2 3^3 when divided by 7 gives a remainder of 6 3^4 when divided by 7 gives a remainder of 4 3^5 when divided by 7 gives a remainder of 5 3^6 when divided by 7 gives a remainder of 1

3^7 when divided by 7 gives a remainder of 3 3^8 when divided by 7 gives a remainder of 2 3^9 when divided by 7 gives a remainder of 6 3^10 when divided by 7 gives a remainder of 4 3^11 when divided by 7 gives a remainder of 5 3^12 when divided by 7 gives a remainder of 1 ..........

The power of 3 i.e 32^32 is equivalent to 2^160 which will be equivalent to remainder when 2^160 is divided 6

2^160 /6 = 2^159 /3 = (3-1)^159 / 3

Now, all terms in (3-1)^159 will be divisible by 3, except for the last term (-1)^159 => -1 => (-1) * 3 +2 . Therefore, the remainder is 2. This, needs to be multiplied by 2, since we had reduced 2^160/6 to 2^159/3. Therefore the final remainder is 4

Therefore, the final value comes out to be 3^4 divided 7, that gives 4 as a remainder.

I got it. Start solving from the top-most power of 32. Just imagine now 2^32 instead of 32^32. So as per cycle, 2^4 = 6. So here 2^32 = 6 as a unit digit. Now, 32^6 means 2^6, which means 2^2 = 4. So the remainder is 4.

Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]

Show Tags

14 Nov 2014, 23:24

Bunuel wrote:

gurpreetsingh wrote:

What is the remainder when \(32^{32^{32}}\) is divided by 7?

A. 5 B. 4 C. 2 D. 0 E. 1

Please do not just post the answer, do explain as well.

I will post the Answer and the explanation after some replies.

If we use the above approach I'd work with prime as a base.

\(32^{{32}^{32}}=(28+4)^{{32}^{32}}\) now if we expand this, all terms but the last one will have 28 as a multiple and thus will be divisible by 7. The last term will be \(4^{{32}^{32}}=4^{{(2^5)}^{32}}=4^{2^{160}}=2^{2^{161}}\). So we should find the remainder when \(2^{2^{161}}\) is divided by 7.

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; ...

The remainder repeats the pattern of 3: 2-4-1.

So we should find \(2^{161}\) (the power of 2) is 1st, 2nd or 3rd number in the above pattern of 3. \(2^{161}\) is 2 in odd power, 2 in odd power gives remainder of 2 when divided by cyclicity number 3, so it's the second number in pattern. Which means that remainder of \(2^{2^{161}}\) divided by 7 would be the same as \(2^2\) divided by 7. \(2^2\) divided by 7 yields remainder of 4.

Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]

Show Tags

09 Jul 2015, 12:39

Bunuel wrote:

gurpreetsingh wrote:

What is the remainder when \(32^{32^{32}}\) is divided by 7?

A. 5 B. 4 C. 2 D. 0 E. 1

Please do not just post the answer, do explain as well.

I will post the Answer and the explanation after some replies.

If we use the above approach I'd work with prime as a base.

\(32^{{32}^{32}}=(28+4)^{{32}^{32}}\) now if we expand this, all terms but the last one will have 28 as a multiple and thus will be divisible by 7. The last term will be \(4^{{32}^{32}}=4^{{(2^5)}^{32}}=4^{2^{160}}=2^{2^{161}}\). So we should find the remainder when \(2^{2^{161}}\) is divided by 7.

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; ...

The remainder repeats the pattern of 3: 2-4-1.

So we should find \(2^{161}\) (the power of 2) is 1st, 2nd or 3rd number in the above pattern of 3. \(2^{161}\) is 2 in odd power, 2 in odd power gives remainder of 2 when divided by cyclicity number 3, so it's the second number in pattern. Which means that remainder of \(2^{2^{161}}\) divided by 7 would be the same as \(2^2\) divided by 7. \(2^2\) divided by 7 yields remainder of 4.

I am lost here Which means that remainder of \(2^{2^{161}}\) divided by 7 would be the same as \(2^2\) divided by 7. \(2^2\) divided by 7 yields remainder of 4.

I cannot figure out why the remainder of \(2^{161}\) equals that of \(2^{2^{161}}\) when the two values are divided by 7.

Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]

Show Tags

19 Jul 2015, 12:15

trueblue wrote:

gurpreetsingh wrote:

What is the remainder when \(32^{32^{32}}\) is divided by 7?

A. 5 B. 4 C. 2 D. 0 E. 1

Please do not just post the answer, do explain as well.

I will post the Answer and the explanation after some replies.

Is the answer B?

Intuitively, i did it like this.

32^32^32 = (28+4)^32^32 As 28 is divisible by 7, we dont need to worry about that part. Hence for the purpose of remainder, our equation boils down to 4^32^32

The cyclicity of 4 is 3 when divided by 7, hence we need to think about the value of 32^32 and what remainder it leaves when divided by 3.

Considering 32^32, it can be broken into (30+2)^32. Again 30^32 is divisible by 3. Hence we need to focus on 2^32. 2^32 can be written as (2*2)^31 = (3+1)^31. As 3^31 is also divisible by 3, we will be left with 1^31. Thus 1 would be the remainder when 32^32 is divided by 3.

This implies that 4 will be the remainder when divided by 7. Hence Answer is B.

Thank you, Karishma. Your explanations always amplify my understanding. Quick question for you - in the last part, i.e.

4*(63+1)^x Gives R4 when divided by 7.

Can you help me understand the binomial part - Binomial dictates that every term except for the last will be divisible by 7. However, can you help me understand why we are not multiplying every term we get with 4 and then trying to find the remainder?

i.e. my working after binomial is: 4*1 / 7 = R4

Why should it not be 4*(term 1/7)*(term 2/7)...*1 / 7 = Remainder

Would really appreciate if you could clarify this portion of my understanding Thank you!!

Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]

Show Tags

11 Feb 2016, 07:56

Bunuel wrote:

If we use the above approach I'd work with prime as a base.

\(32^{{32}^{32}}=(28+4)^{{32}^{32}}\) now if we expand this, all terms but the last one will have 28 as a multiple and thus will be divisible by 7. The last term will be \(4^{{32}^{32}}=4^{{(2^5)}^{32}}=4^{2^{160}}=2^{2^{161}}\). So we should find the remainder when \(2^{2^{161}}\) is divided by 7.

Hi Bunuel,

I fail to understand how \(4^{2^{160}}\) is equal to \(2^{2^{161}}\) Could you please shed some light on it?

Thank you!

gmatclubot

Re: What is the remainder when 32^32^32 is divided by 7?
[#permalink]
11 Feb 2016, 07:56

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

There is without a doubt a stereotype for recent MBA grads – folks who are ambitious, smart, hard-working, but oftentimes lack experience or domain knowledge. Looking around and at...