What is the remainder when 32^32^32 is divided by 7? : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 21 Jan 2017, 17:18

### 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: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 226

Kudos [?]: 1620 [7] , given: 235

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

### Show Tags

02 Sep 2010, 17:23
7
KUDOS
34
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

45% (02:04) correct 55% (01:22) wrong based on 1023 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA

_________________

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

Last edited by gurpreetsingh on 04 Sep 2010, 10:43, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 36590
Followers: 7092

Kudos [?]: 93364 [12] , given: 10557

### Show Tags

04 Sep 2010, 10:44
12
KUDOS
Expert's post
23
This post was
BOOKMARKED
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.
_________________
Intern
Joined: 02 Sep 2010
Posts: 3
Followers: 0

Kudos [?]: 8 [5] , given: 1

### Show Tags

03 Sep 2010, 02:59
5
KUDOS
2
This post was
BOOKMARKED
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.
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 226

Kudos [?]: 1620 [3] , given: 235

### Show Tags

04 Sep 2010, 10:26
3
KUDOS
6
This post was
BOOKMARKED
gurpreetsingh wrote:
What is the remainder when $$32^{32^{32}}$$ is divided by 7?

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

Many of you got the correct answer but wrong explanation and it is the biggest losing point of the learning as the same question is never going to come.

trueblue nailed it with correct explanation.

Here is my explanation and I would request the Quant guru of Gmat Club - Bunnel to correct and improve my explanation for the solutions.

$$32^{32^{32}}$$ looks quite daunting, but remember one thing on exams like GMAT you will always get tricky and convoluted questions. But they can solved by simple Quant basics.

Always reduce your question.... Rof means remainder of

Rof $$32^{32^{32}}$$ when divided by 7 = Rof $$4^{32^{32}}$$ by 7 as 28+4 = 32

To calculate Rof $$4^{32^{32}}$$ , we need to understand how to reduce it further.

Whenever we are solving remainder questions we always reduce it to the minimum value. To reduce the powers of 4 we need to find $$4^x$$ such that when $$4^x$$ is divided by 7 the remainder is either 1 or -1

Reason : Rof $$4^{xy}$$ = Rof $$4^x * 4^y$$ = Rof $$4^{x}$$ * Rof $$4^{y}$$

If remainder of Rof $$4^{x}$$ =1 , then we can reduce the Rof $$4^{y}$$ multiple times until y>x.

Since Rof $$4^{3}$$ when divided by 7 is 1, if we can reduce $$4^{32^{32}}$$ to the form of $$4^{3k+r}$$ we can easily eliminate redundant powers of 4.

To represent $$4^{32^{32}}$$ as $$4^{3k+r}$$ we need to represent $$32^{32}$$ in the form of $$3k+r$$.

Now to represent $$32^32$$ in the form of $$3k+r$$, we have to find the remainder when $$32^{32}$$ is divided by 3. That will give the value of r.

Rof $$32^{32}$$ when divided by 3 = Rof $$2^{32}$$ = Rof $$2^{4*{8}}$$

Since Rof $$2^4$$when divided by 3 = 1, => Rof $$2^{4*{8}}$$ = 1

Hence r = 1 => $$32^{32}$$ = 3k+1

Now coming back to the main question.

Rof $$4^{32^{32}}$$ = Rof $$4^{3k+1}$$ = Rof $$4^{3k}$$ * Rof $$4^1$$ = 1*4 = 4

Hence B.

Whenever you see such question, always apply the above rule.
_________________

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

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7125
Location: Pune, India
Followers: 2137

Kudos [?]: 13681 [3] , given: 222

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

### Show Tags

20 Apr 2012, 08:54
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

Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 408
Location: Milky way
Schools: ISB, Tepper - CMU, Chicago Booth, LSB
Followers: 8

Kudos [?]: 196 [2] , given: 50

### Show Tags

02 Sep 2010, 18:16
2
KUDOS
2
This post was
BOOKMARKED
My approach:

32^32^32 -- $$2^5$$^$$2^5$$^$$2^5$$

=> 2^800

$$(2^x)/7$$ has a cyclicity or repeatability of 3.
That is $$(2^1)/7 - 2$$, $$(2^2)/7- 4$$, $$(2^3)/7 - 1$$ .....

Hence (2^800)/7 boils down to the same as $$(2^2)/7$$ which is 4.

_________________

Support GMAT Club by putting a GMAT Club badge on your blog

Last edited by ezhilkumarank on 02 Sep 2010, 18:32, edited 1 time in total.
Verbal Forum Moderator
Joined: 02 Aug 2009
Posts: 4233
Followers: 328

Kudos [?]: 3458 [2] , given: 101

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

### Show Tags

16 Feb 2016, 06:57
2
KUDOS
Expert's post
nishantdoshi wrote:
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.

4^{{32}^{32}}=4^{{(2^5)}^{32}}=4^{2^{160}}=2^{2^{161}}

how does it change to 161?????

Hi,
I'll try to explain to you..
$$4^{2^{160}}= {2^{2*2^{160}}={2^{2^{160+1}}=2^{2^{161}}$$

Hope it helps
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Manager
Status: He who asks is a fool for five minutes, but he who does not ask remains a fool forever
Joined: 20 Aug 2010
Posts: 101
Followers: 2

Kudos [?]: 7 [1] , given: 2

### Show Tags

02 Sep 2010, 19:29
1
KUDOS
1
This post was
BOOKMARKED
Is it 4? My took following approach,

Last digit of 32^32 will be 4 , so last digit of 32^4 will be same as 32^1. so 32/7 =4.

Please correct me if I am wrong
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 226

Kudos [?]: 1620 [1] , given: 235

### Show Tags

03 Sep 2010, 05:33
1
KUDOS
Is this even a GMAT Question...??????
because if it is I need to start worrying about Quants as well....

If this question is asked in the exam, it will be at 750+ level. So do not worry even if you are not able to solve this. Sometimes we learn concepts by understanding the explanation of difficult problem and use them while solving similar questions.

Give it a try and do not lose your confidence. When I will post the explanation do read and understand it carefully.
_________________

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 [?]: 34 [1] , given: 0

### Show Tags

03 Sep 2010, 06:51
1
KUDOS
1
This post was
BOOKMARKED
32^32^32 % 7 = ?

I am using Euler's method, search on wikipedia if you need proof, else try to follow the steps :

HCF(32,7) = 1
"phi" 7 = 6 (it is the number of positive integers less than 7 and prime to 7.. In fact for any prime number "n", it will be "n-1").

=> 32^6 mod 7 = 1 (mod is same as "%")

(To make sure you understand it, please try for any number n!=7, n^6 mod 7 = 1)

So, we now need to express, 32^32 = 6x+k

i.e. 32^32 % 6 = ?

To make it easier, lets try to find out 16^32 % 3 and multiply the remainder by 4 (since 32 and 6 has a common factor 2, and also it is easier/helpful to get a remainder divided by a prime number)

Apply the same approach as shown above :
HCF(16,3) = 1
"phi" 3 = 2

=> 16^2 mod 3 = 1
=> 16^32 mod 3 = 1 => 32^32 mod 6 = 4*1 = 4

So, 32^32 mod 6 = 6y+4

Therefore;
32^32^32 mod 7 = 32^(6y+4) mod 7 = 32^4 mod 7 = (28+4)^4 mod 7 = 4^4 mod 7 = 4

(Hopefully, I didn't make any typo.... Let me know if there is any problem with understanding this)

Thanks

Last edited by anshumishra on 03 Sep 2010, 09:18, edited 1 time in total.
Intern
Joined: 05 Jun 2015
Posts: 26
Location: Viet Nam
GMAT 1: 740 Q49 V41
GPA: 3.66
Followers: 0

Kudos [?]: 6 [1] , given: 444

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

### Show Tags

11 Feb 2016, 08:00
1
KUDOS
2
This post was
BOOKMARKED
VeritasPrepKarishma wrote:

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.

Hi Karishma,

Is it always better to avoid subtraction when using binomial theorem? Here is my approach, and I found that if there is a negative sign in the equation, things can get a little bit complicated.

$$32^{32^{32}}$$ = $$(35-3)^{32^{32}}$$

Now we only care about $$(-3)^{32^{32}} = 3^{32^{32}}$$

We know that 27 is 1 less than a multiple of 1 and that $$3^{3} = 27$$

$$32^{32} = (33-1)^{32}$$
Hence $$3^{32^{32}} = 3^{3x+1} = 3*3^{3x} = 3*27^{x} = 3*(28-1)^x$$

Here we have to check whether x is even or not in order to decide the value of the remainder (whether it is 3 or 4).
One additional step: Since 32 is even, $$32^{32}$$ is even. $$32^{32} = 3x+1$$ therefore $$x$$ is odd.

We now can conclude that r = -3. Therefore r = 4.

My question is: 'Is it always safer to avoid subtraction when it comes to binomial theorem?'
Thank you Karishma!
Verbal Forum Moderator
Joined: 02 Aug 2009
Posts: 4233
Followers: 328

Kudos [?]: 3458 [1] , given: 101

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

### Show Tags

17 Mar 2016, 01:07
1
KUDOS
Expert's post
Chiragjordan wrote:
Looked ...Re looked ...RE RE looked...
NO where to go
chetan2u you gotta help with this..
Regards

Hi,

Before I explain you this Q, I will just explain the Binomial theorem--

$$(a + b)^n = nC0 a^n + nC1 a^{n − 1}b + nC2 a^{n − 2}b^2 + nC3 a^{n − 3}b^3+.....nC(n-1)b^{n − 1}a + nC0 b^n$$
so if you look at this whatever be the power 'n' be, its all terms will be div by a, except one term nC0 b^n, same is the case with div by b..

EXAMPLE--
$$(x+5)^4 = x^4 + 20x^3 + 150x^2 + 500x + 625$$
here all terms except 625 will be div by x and all terms except x^4 wil be div by 5..

lets use this to hep in finding remainders..

When $${{32}^{32}}^{32}$$ is div by 7..
we know 8=7+1..
lets convert in this form..

$${{2^5}^{32}}^{32}$$=$${{(2^3*2^2)}^{32}}^{32}$$=$${{2^3}^{32}}^{32}*{{2^2}^{32}}^{32}$$..

now we have two terms
1)$${{2^3}^{32}}^{32}$$=$${{8}^{32}}^{32}={{7+1}^{32}}^{32}$$
so here all terms in the expansion will be div by 7 except the one with only 1s, that is $$nCn7^0*{1^{32}}^{32}$$
So remainder will be 1 for this..

2)$${{2^2}^{32}}^{32}$$..
similarly make the power inside as 2^3 instead of 2^2 and work ahead..
Or
4^1=4 it will leave the remainder of 4
4^2=16, this will leave 2
4^3=64, this will leave 1

so get above in this form
$${{2^2}^{32}}^{32}$$.= $${{4}^{30}}^{32}$$*$${{4}^{2}}^{32}$$.

again $${{4}^{30}}^{32}$$ will leave remainder 1, and $${{4}^{2}}^{32}$$ will leave 2^32 as remainder

$$2^{32}= 2^{30}*2^2$$..
$$2^{30} = 8^{10}$$, so here too remainder is 1..
2^2 will leave 4 as remainder

ans 4

You may find this lengthy, since I have explained all terms for your understanding..
but If you get hang of it, it will be easier and faster..
Ofcourse there are always shortcuts as per each Q but we should know the standard method and WHY/ of each Q.

_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Math Expert
Joined: 02 Sep 2009
Posts: 36590
Followers: 7092

Kudos [?]: 93364 [1] , given: 10557

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

### Show Tags

20 Oct 2016, 22:28
1
KUDOS
Expert's post
ac556 wrote:
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.

Hope it's clear.

Hello Bunuel, I am having trouble understanding this bit - $$2^{161}$$ 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.

Will it be possible for you to explain this in a bit more detail? Many thanks.

2^1 = 2 divided by 3 --> remainder 2;
2^3 = 8 divided by 3 --> remainder 2;
2^5 = 32 divided by 3 --> remainder 2;
...
_________________
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 226

Kudos [?]: 1620 [0], given: 235

### Show Tags

02 Sep 2010, 18:22
How you got 2^800? read the question carefully.

Just quote my question and see how I have used the mathematical operation for the powers. This way you can represent them better as your solution is quite difficult to understand.
_________________

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

Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 408
Location: Milky way
Schools: ISB, Tepper - CMU, Chicago Booth, LSB
Followers: 8

Kudos [?]: 196 [0], given: 50

### Show Tags

02 Sep 2010, 18:47
gurpreetsingh wrote:
How you got 2^800? read the question carefully.

Just quote my question and see how I have used the mathematical operation for the powers. This way you can represent them better as your solution is quite difficult to understand.

Got it. Here is the updated post.

$$32^{{32}^{32}$$ -- $${{2^5}^{{2^5}^{2^5}}}$$

$${{2^5}^{2^5}}$$ is $${2^{160}}$$.

Again $${{2^{160}}^{32}}$$ is $${2^{5120}}$$

(2^x)/7 has a cyclicity or repeatability of 3.
That is $${2^1}/7 - 2$$, $${2^2}/7- 4$$, $${2^3}/7 - 1$$ .....

Hence $${2^{5120}}/7$$ boils down to the same as $${2^{2}}/7$$ which is 4.

_________________

Support GMAT Club by putting a GMAT Club badge on your blog

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 226

Kudos [?]: 1620 [0], given: 235

### Show Tags

02 Sep 2010, 19:05
ezhilkumarank wrote:
gurpreetsingh wrote:
How you got 2^800? read the question carefully.

Just quote my question and see how I have used the mathematical operation for the powers. This way you can represent them better as your solution is quite difficult to understand.

Got it. Here is the updated post.

$$32^{{32}^{32}$$ -- $${{2^5}^{{2^5}^{2^5}}}$$

$${{2^5}^{2^5}}$$ is $${2^{160}}$$.

Again $${{2^{160}}^{32}}$$ is $${2^{5120}}$$

(2^x)/7 has a cyclicity or repeatability of 3.
That is $${2^1}/7 - 2$$, $${2^2}/7- 4$$, $${2^3}/7 - 1$$ .....

Hence $${2^{5120}}/7$$ boils down to the same as $${2^{2}}/7$$ which is 4.

Take my advice when ever you solve your question always check the scope and domain.

do you think $$32^{32^{32}}$$ can be equal to $${2^{5120}}$$ ?

$$32^{32^{32}}$$ = $$2^{5*{32^{32}}}$$

$$2^{5*{32^{32}}}$$ should be equal to $${2^{5120}}$$
=> $$5*{32^{32}}$$ should be equal to 5120. ? This is wrong.

Just cross check always whether you are making right moves or not.
I hope you know your mistake now?
_________________

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: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 226

Kudos [?]: 1620 [0], given: 235

### Show Tags

02 Sep 2010, 19:45
PositiveSoul wrote:
Is it 4? My took following approach,

Last digit of 32^32 will be 4 , so last digit of 32^4 will be same as 32^1. so 32/7 =4.

Please correct me if I am wrong

Last two digits of 32^32 is 76 not 4.

Also last digit of 32^4 is 6 and it is not same as 32^1
_________________

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

Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 408
Location: Milky way
Schools: ISB, Tepper - CMU, Chicago Booth, LSB
Followers: 8

Kudos [?]: 196 [0], given: 50

### Show Tags

02 Sep 2010, 20:19
gurpreetsingh wrote:
ezhilkumarank wrote:
gurpreetsingh wrote:
How you got 2^800? read the question carefully.

Just quote my question and see how I have used the mathematical operation for the powers. This way you can represent them better as your solution is quite difficult to understand.

Got it. Here is the updated post.

$$32^{{32}^{32}$$ -- $${{2^5}^{{2^5}^{2^5}}}$$

$${{2^5}^{2^5}}$$ is $${2^{160}}$$.

Again $${{2^{160}}^{32}}$$ is $${2^{5120}}$$

(2^x)/7 has a cyclicity or repeatability of 3.
That is $${2^1}/7 - 2$$, $${2^2}/7- 4$$, $${2^3}/7 - 1$$ .....

Hence $${2^{5120}}/7$$ boils down to the same as $${2^{2}}/7$$ which is 4.

Take my advice when ever you solve your question always check the scope and domain.

do you think $$32^{32^{32}}$$ can be equal to $${2^{5120}}$$ ?

$$32^{32^{32}}$$ = $$2^{5*{32^{32}}}$$

$$2^{5*{32^{32}}}$$ should be equal to $${2^{5120}}$$
=> $$5*{32^{32}}$$ should be equal to 5120. ? This is wrong.

Just cross check always whether you are making right moves or not.
I hope you know your mistake now?

Not sure if I got what you mentioned, but upon analyzing my approach I found the error.

$${{2^5}^{2^5}} is {2^{160}}$$. -- This is correct.

[highlight]Again $${{2^{160}}^{32}} is {2^{5120}}$$[/highlight] -- This is the incorrect part.

The above line should have been $${32^{2^{160}}}$$.

Now if I understand correctly, this was the only error in my approach. Further proceeding to solve the question....

{32^1}/7 - 4, {32^2}/7 - 2, {32^3}/7 - 1, {32^4}/7 - 4 .... hence the repeatability is 3.

Now $$32^{2^{160}}$$ should be the same as the power of 32 which is $${2^{160}}$$ divided by 3.

$${2^{160}}$$ divided by 3 is the same as $${2^{1}}$$ divided by 3 -- which is 2.

Now combining them $$32^{2^{160}}/7$$ is the same as $${32^{2^1}}/7$$

which is 2. Final answer should be 2 (C).

Hope I have answered correctly this time. I will await the final solution and the explanation.
_________________

Support GMAT Club by putting a GMAT Club badge on your blog

Manager
Joined: 17 Oct 2008
Posts: 196
Followers: 1

Kudos [?]: 25 [0], given: 11

### Show Tags

02 Sep 2010, 23:03
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
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 226

Kudos [?]: 1620 [0], given: 235

### Show Tags

03 Sep 2010, 05:31
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

Check this : tough-remainder-question-100316.html#p773920

You are making the same mistake. Also 1024 - 32^2 not 32^32.
_________________

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

Re: Tough remainder question   [#permalink] 03 Sep 2010, 05:31

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

Similar topics Replies Last post
Similar
Topics:
2 What is the remainder when 7^442 is divided by 10? 5 22 Sep 2016, 16:29
If n divided by 7 has a remainder of 2, what is the remainder when 3 5 21 Mar 2016, 06:28
29 What is the remainder when 333^222 is divided by 7? 19 21 Jul 2013, 01:16
1 What is the remainder when 7^381 is divided by 5 ? 5 06 Oct 2009, 00:30
15 What is the remainder when 7^74 - 5^74 is divided by 24? 13 03 Jul 2008, 12:19
Display posts from previous: Sort by