GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 06 Jul 2020, 03:01 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # What is the remainder when 32^32^32 is divided by 7?

Author Message
TAGS:

### Hide Tags

SVP  Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2448
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

1
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 Support GMAT Club by putting a GMAT Club badge on your blog/Facebook 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
SVP  Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2448
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

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 Support GMAT Club by putting a GMAT Club badge on your blog/Facebook 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: 71
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

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: 71
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

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 SVP  Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2448
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

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 Support GMAT Club by putting a GMAT Club badge on your blog/Facebook 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: 19
Location: Texas
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

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: 870
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

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: 07 Dec 2011
Posts: 63
Location: India
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

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: 8
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

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 V
Joined: 16 Oct 2010
Posts: 10634
Location: Pune, India
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

5
2
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

SVP  Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2448
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

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 Support GMAT Club by putting a GMAT Club badge on your blog/Facebook 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
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

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: 95
Location: India
Concentration: Strategy, Finance
WE: Engineering (Energy and Utilities)
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

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 2008
Posts: 57
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

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
Intern  Joined: 04 Jan 2014
Posts: 3
Concentration: Finance, Strategy
GMAT Date: 07-31-2014
GPA: 3.95
WE: Engineering (Energy and Utilities)
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

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

Consider the following:

when (32)^1, unit digit =2.
when (32)^2, unit digit = 4.
when (32)^3, unit digit = 8.
when (32)^4, unit digit = 6.
when (32)^5, unit digit = 2.

Hence (32)^x, where x is an integer has a cyclicity of 4.

=> (32)^32 will have 2 as unit digit; this is because (32/4) = 8. Hence, original expression becomes:

{(32)^2}/7 = ?

From above, when (32)^2, unit digit = 4.

Expression becomes 4/7 which has a remainder of 4.

Took me 1:15 mins to solve.
CrackVerbal Representative G
Affiliations: CrackVerbal
Joined: 03 Oct 2013
Posts: 1876
Location: India
GMAT 1: 780 Q51 V46
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

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

2^5(32)(32)

2^1/7 = Remainder = 2
2^2 /7 = Remainder = 4
2 ^3 /7 = Remainder= 1
16/7 = Remainder = 2
32/7 = Remainder 4
the process continues

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.
_________________
Intern  Joined: 03 Apr 2014
Posts: 1
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

ramana 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 got B

$$32^{32^{32}}$$ can be reduced as {32^32} ^ 32 = 1024 ^ 32

and 1024 = 2^10

= (2)^10*32 -> 2^320

R[ (2^x)/7 ] is cyclical,2^320/7 is same as 2^2/7 and the answer is 4!

correct me if am wrong

$$32^{32^{32}} = 32^{1.461501637331 *10^{48}}$$

You do top down when there is no parentheses and so you do 32^32 and than you take 32 to the power of what you got for 32^32
Manager  Joined: 04 Jan 2014
Posts: 93
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

gurpreetsingh wrote:
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

Is the remainder 5?

9 x 3 + 5 =32
Intern  Joined: 27 Sep 2013
Posts: 11
Location: Netherlands
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

Found this in .30 sec, hope I'm right though.

I (In the last part - 32^32^32) find the cyclicity of 32 (or, since it ends with a 2, find the cyclicity of 2)
2...4...8...6 .

II We can see that 32 is evenly divisible by 4 so the units digit of 32^32^32 is something with a 2.

III Squaring 32 (32^2) ends with a units digit of 4

IV dividing 4 by 7 leaves a remainder of 4.

Hence, B.

Math Expert V
Joined: 02 Sep 2009
Posts: 64979
Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

### Show Tags

satsymbol wrote:
is this a GMAT type question?

Concepts tested are relevant for the GMAT, though question itself is harder than one can expect on the real test.
_________________ Re: What is the remainder when 32^32^32 is divided by 7?   [#permalink] 08 May 2014, 07:14

Go to page   Previous    1   2   3   4    Next  [ 66 posts ]

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