Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
GMAT Club

 It is currently 29 Mar 2017, 08:38

# LIVE NOW:

### 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

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

Track

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# What is the remainder when 32^32^32 is divided by 7?

Author Message
TAGS:

### Hide Tags

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2793
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 230

Kudos [?]: 1669 [0], given: 235

### Show Tags

04 Sep 2010, 14:05
1
This post was
BOOKMARKED
anshumishra wrote:
32^32^32 % 9 = ?

32^32^32 = 2^2^161
Here the remainder repeats the pattern of 6: 2,4,8,7,5,1

So, 2^2^161 % 9 = 2^5 % 9 = 5

How you got the step in red? Do not copy the Bunnel's explanation, this time it has to be divided by 9 not 7.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2793
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 230

Kudos [?]: 1669 [0], given: 235

### Show Tags

04 Sep 2010, 14:12
Guys bear with me , I m posting the question created by me.

What is the remainder when $$11^{11^{11^{11}}.....10 times}$$ is divided by 4.

a. 1
b. 0
c. 3
d. 2
e. None

This is a very easy question. This concept will help in many other questions. THINK LOGICALLY AND REVISE BASICS OF REMAINDERs.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Manager
Joined: 25 Jun 2010
Posts: 91
Followers: 1

Kudos [?]: 37 [0], given: 0

### Show Tags

04 Sep 2010, 15:03
gurpreetsingh wrote:
Guys bear with me , I m posting the question created by me.

What is the remainder when $$11^{11^{11^{11}}.....10 times}$$ is divided by 4.

a. 1
b. 0
c. 3
d. 2
e. None

This is a very easy question. This concept will help in many other questions. THINK LOGICALLY AND REVISE BASICS OF REMAINDERs.

11^z %4 = (12-1)^z %4 = (-1)^z % 4 = 3 if z is odd, else 1 when z is even
Manager
Joined: 25 Jun 2010
Posts: 91
Followers: 1

Kudos [?]: 37 [0], given: 0

### Show Tags

04 Sep 2010, 15:07
gurpreetsingh wrote:
anshumishra wrote:
32^32^32 % 9 = ?

32^32^32 = 2^2^161
Here the remainder repeats the pattern of 6: 2,4,8,7,5,1

So, 2^2^161 % 9 = 2^5 % 9 = 5

How you got the step in red? Do not copy the Bunnel's explanation, this time it has to be divided by 9 not 7.

Yeah, thought to copy the partial solution of Bunnel to save sometime, however made mistake because of rushing through it :
32^32^32 %9 = (27+5)^32^32 % 9 = 5^32^32 % 9 = 5 ^ 2^160 % 9

The cyclicity here is 6 , so it could be solved the same ways.
I am not going to try it again this time
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2793
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 230

Kudos [?]: 1669 [0], given: 235

### Show Tags

04 Sep 2010, 15:39
anshumishra wrote:
gurpreetsingh wrote:
anshumishra wrote:
32^32^32 % 9 = ?

32^32^32 = 2^2^161
Here the remainder repeats the pattern of 6: 2,4,8,7,5,1

So, 2^2^161 % 9 = 2^5 % 9 = 5

How you got the step in red? Do not copy the Bunnel's explanation, this time it has to be divided by 9 not 7.

Yeah, thought to copy the partial solution of Bunnel to save sometime, however made mistake because of rushing through it :
32^32^32 %9 = (27+5)^32^32 % 9 = 5^32^32 % 9 = 5 ^ 2^160 % 9

The cyclicity here is 6 , so it could be solved the same ways.
I am not going to try it again this time

The most important thing is to learn the concept.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Intern
Joined: 10 Oct 2010
Posts: 23
Location: Texas
Followers: 3

Kudos [?]: 18 [0], given: 1

### Show Tags

10 Oct 2010, 03:09
Hi,

I'm new here. First, I wanted to say thank you to everyone for all of the awesome questions and explanations throughout the forum.

Second, here are my explanations for the three questions that have been posted.

R{x/y} represents remainder of x divided by y.
R{(ab)/y} = R{ (R{a/y}*R{b/y}) / y} <---- I found this on one of the forum posts.
Therefore, R{(a^c)/y} = R{(R{a/y}^c) / y} <---- I used a nested version of this on all three problems.

Problem 1)
R{32^(32^32)/7}
= R{(R{(R{32/7}^32) / 7}^32) / 7} <-------- R{32/7} = 4
= R{(R{( 4 ^32) / 7}^32) / 7} <-------- R{(4^32)/7} = 2, (R cycles 4,2,1,4,2,1...)
= R{( 2 ^32) / 7} <-------- R{(2^32)/7} = 4, (R cycles, 2,4,1,2,4,1...)
= 4

Problem 2)
R{32^(32^32)/9}
= R{(R{(R{32/9}^32) / 9}^32) / 9} <-------- R{32/9} = 5
= R{(R{( 5 ^32) / 9}^32) / 9} <-------- R{(5^32)/9} = 7, (R cycles 5,7,8,4,2,1,5...)
= R{( 7 ^32) / 9} <-------- R{(7^32)/9} = 4, (R cycles, 7,4,1,7,4,1...)
= 4

Problem 3)
R{11^(11^(11^(11^(11...etc))))/4}
= R{(R{(R{11/4}^11) / 4}^11....etc.) / 4} <-------- R{11/4} = 3
= R{(R{( 3 ^11) / 3}^11....etc.) / 4} <-------- R{(3^11)/4} = 3, (R cycles 3,1,3,1...)
= R{( 3 ^11....etc.) / 4} <-------- R{(3^11)/4} = 3, (R cycles, 3,1,3,1...)
= 3
Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1708
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Followers: 102

Kudos [?]: 959 [0], given: 109

### Show Tags

17 Oct 2010, 16:25
Bunuel wrote:
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.

Hi Bunuel,
I don't understand the part of the explanation highlighted in red.
Before that part, you analyzed $$2^{161}$$ and concluded that its remainder is 4 (second number in pattern). I am Ok with that.
However, I don't understand when you conclude that the remainder will be also 4 when you analyze $$2^{2^{161}}$$. I don't follow you.
Thanks!
_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

GMAT Club Premium Membership - big benefits and savings

Manager
Joined: 25 Jul 2010
Posts: 175
WE 1: 4 years Software Product Development
WE 2: 3 years ERP Consulting
Followers: 7

Kudos [?]: 51 [0], given: 15

### Show Tags

17 Oct 2010, 19:12
Was trying to use remainder theorem but just gave up :D
_________________

Director
Joined: 23 Apr 2010
Posts: 584
Followers: 2

Kudos [?]: 83 [0], given: 7

Find the remainder when 32^32^32 is divided by 7? [#permalink]

### Show Tags

28 Sep 2011, 01:47
Find the remainder when 32^32^32 is divided by 7?

I know this question has been raised several times on this forum, but I can't find the post.

Thanks.
Intern
Joined: 05 Jul 2011
Posts: 25
Followers: 0

Kudos [?]: 0 [0], given: 2

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

### Show Tags

28 Sep 2011, 22:33
Do you mean 32^{32^{32}} ? If yes, the answer is here
tough-remainder-question-100316-20.html
Director
Joined: 23 Apr 2010
Posts: 584
Followers: 2

Kudos [?]: 83 [0], given: 7

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

### Show Tags

29 Sep 2011, 02:14
nrgmat, thanks a lot. That's what I've been looking for.
Manager
Joined: 18 Jun 2010
Posts: 148
Followers: 0

Kudos [?]: 35 [0], given: 2

### Show Tags

19 Oct 2011, 21:35
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

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.

Similar problem: remainder-99724.html?hilit=expand%20this,%20all%20terms#p768816

Hope it's clear.

A new concept learnt. Thanks
Intern
Joined: 01 Oct 2011
Posts: 24
GMAT 1: 730 Q49 V41
WE: Information Technology (Computer Software)
Followers: 1

Kudos [?]: 15 [0], given: 21

### Show Tags

19 Oct 2011, 23:25
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

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

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.

Do let me know if i am wrong in my thinking.

Thanks.

Very good explanation... thanks
_________________

- Success is not final, failure is not fatal: it is the courage to continue that counts

http://gmatclub.com/forum/finally-gmat-is-over-730-49q-41v-130632.html

Manager
Joined: 07 Dec 2011
Posts: 174
Location: India
Followers: 1

Kudos [?]: 41 [0], given: 31

### Show Tags

23 Mar 2012, 05:35
find cycle of remainders which are 2,4,1
total power of 2 = 32x32x5 = 5120 divide taht by 3 and get remainder of 2 so the second value in the cycle ie (4) is the answer. B.
Intern
Joined: 12 Mar 2012
Posts: 16
Followers: 0

Kudos [?]: 7 [0], given: 19

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

### Show Tags

19 Apr 2012, 10:51
gurpreetsingh wrote:
What is the remainder when $$32^{32^{32}}$$ is divided by 7?

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

Check the solution here : tough-remainder-question-100316.html#p774893

Guys, found one more method to solve the problem

32^32^32

=(28+4)^32^32

28^32^32 is divisible by seven and we are only concerned about 4^32^32= 4^2^160= 2^2^161

Now 2^161= 2^10^6 X 2^1
= 1024^6X2^1

last digit of 1024^6 will be last didgit of 4^6 i.e 2^12 i.e 2^10X 2^2 i.e 1024X4 hence last digit of 1024^6 will be( 4X4 =16 ) 6

last digit of 1024^6X2=> 6X2=12 => 2
thus equation boils down to 2^2= 4 divided by 7 , reminder will be 4 the ans.

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7256
Location: Pune, India
Followers: 2205

Kudos [?]: 14370 [3] , given: 222

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

### Show Tags

20 Apr 2012, 09:54
3
KUDOS
Expert's post
1
This post was
BOOKMARKED
Cmplkj123 wrote:

Now 2^161= 2^10^6 X 2^1
= 1024^6X2^1

Definitely a good effort but how did you get this?

This is how I would do it using binomial theorem:

$$32^{32^{32}}$$ divided by 7

$$(28 + 4)^{32^{32}}$$ divided by 7

We need to figure out $$4^{32^{32}}$$ divided by 7
We know that 64 is 1 more than a multiple of 7 and that $$4^3 = 64$$

But how many 3s do we have in $$32^{32}$$?
$$32^{32} = (33 - 1)^{32}$$ so when we divide $$32^{32}$$ by 3, we get a remainder of 1.

$$4^{32^{32}} = 4^{3x+1} = 4*4^{3x} = 4*64^x = 4*(63+1)^x$$

When this is divided by 7, remainder is 4.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2793
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 230

Kudos [?]: 1669 [0], given: 235

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

### Show Tags

28 Apr 2012, 15:37
Thanks Karishma for simplifying the solution.
Those who need detailed solution for any other combination then follow
what-is-the-remainder-when-32-32-32-is-divided-by-100316.html#p774893
to understand the basics.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Intern
Joined: 13 Jul 2012
Posts: 4
Followers: 0

Kudos [?]: 1 [0], given: 0

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

### Show Tags

14 Jul 2012, 09:27
Thanks so much for 2 excellent ways of solution of gurpreetsingh and Bunuel. I want to do some practice.

Similar question to test what you have learnt from the previous post.
What is the remainder when 32^{32^{32}} is divided by 9?
A. 7
B. 4
C. 2
D. 0
E. 1

Firstly, I want to try gurpreetsingh's way.
The remainder of 32^{32^{32}} when divided by 9 is equal to the remainder of 5^{32^{32}} dividing by same number (32=9x+5).
Obviously, when dividing 5^{6} = 15,625 by 9 we have the remainder of 1.
So, 5^{32^{32}} can be rewritten as a product of k times of 5^{6} and others: 5^{32^{32}} = 5^{6}*5^{6}*...*5^{6}+5^{r}
We have to know the value of r, which is the remainder when dividing 32^{32} by 6 or 32^{32} = 6k+r

Similarly, the remainder of 32^{32} when divided by 6 is equal to the remainder of 2^{32} dividing by same number (32=6y+2).
2^{32} = 2^{10}*2^{10}*2^{10}*2^{2}
Because 2^{10} = 1024 = 6z+4, the remainder when 2^{32} is divided by 6 must be the remainder when dividing (4*4*4*4) by 6.
4^{4} = 256 = 6m+4

So we have r=4, and the remainder we need to find out is 4 (5^{4} = 625 = 9n+4)

Next, move to another solution by Bunuel.
We can do the same logic to reach for the remainder of 5^{32^{32}} dividing by 9.
Also, we can find the same pattern for 9:
5^1 divided by 9 yields remainder of 5;
5^2 divided by 9 yields remainder of 7;
5^3 divided by 9 yields remainder of 8;
5^4 divided by 9 yields remainder of 4;
5^5 divided by 9 yields remainder of 2;
5^6 divided by 9 yields remainder of 1;

5^7 divided by 9 yields remainder of 5;
5^8 divided by 9 yields remainder of 7;
5^9 divided by 9 yields remainder of 8;
5^10 divided by 9 yields remainder of 4;
5^11 divided by 9 yields remainder of 2;
5^12 divided by 9 yields remainder of 1;
...

The remainder repeats the pattern of 6: 5-7-8-4-2-1.
So we have to rewrite 32^{32} in another way: 32^{32} = 6k+r and find out r to determine which is the correct remainder in the pattern.
Similar logic, we can easily find that r = 4, respectively, the remainder should be at the 4th palce in the pattern, it should be 4.

So both ways are similar to each other.
Very interesting!
Manager
Joined: 28 Dec 2012
Posts: 114
Location: India
Concentration: Strategy, Finance
GMAT 1: Q V
WE: Engineering (Energy and Utilities)
Followers: 3

Kudos [?]: 69 [0], given: 90

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

### Show Tags

31 Jan 2013, 01:28
each 32 in the product gives remainder 4 with 7. So remainder is 4^32^32
now lets divide powers of 4 with 7
we get remainder as 4,2,1,4 ... with 4^1, 4^2, 4^3, 4^4 respectively
thus remainder repeats with a cycle of 3.
Lets divide 32^32 with this cycle of 3
32^32 = (33-1)^32 which on division by 3 gives remainder 1 (-1^32)

that means 4^32^32 = 4^(3k + 1) since cylce is 3 on division of powers of 4 by 7, the remainder is 4^1 = 4 Answer.

KUDOS if u like
_________________

Impossibility is a relative concept!!

Manager
Joined: 14 Nov 2011
Posts: 152
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)
Followers: 0

Kudos [?]: 17 [0], given: 103

### Show Tags

08 Jun 2013, 09:29
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

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.

Similar problem: remainder-99724.html?hilit=expand%20this,%20all%20terms#p768816

Hope it's clear.

Hi Bunnel,

Is this a GMAT question ?
Re: Tough remainder question   [#permalink] 08 Jun 2013, 09:29

Go to page   Previous    1   2   3   4    Next  [ 72 posts ]

Similar topics Replies Last post
Similar
Topics:
3 What is the remainder when 7^442 is divided by 10? 5 22 Sep 2016, 17:29
If n divided by 7 has a remainder of 2, what is the remainder when 3 5 21 Mar 2016, 07:28
31 What is the remainder when 333^222 is divided by 7? 19 21 Jul 2013, 02:16
47 What is the remainder when (18^22)^10 is divided by 7 ? 28 24 Aug 2010, 02:35
1 What is the remainder when 7^381 is divided by 5 ? 5 06 Oct 2009, 01:30
Display posts from previous: Sort by