It is currently 21 Sep 2017, 03:49

### 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

Manager
Joined: 22 Sep 2012
Posts: 141

Kudos [?]: 73 [0], given: 49

Concentration: Strategy, Technology
WE: Information Technology (Computer Software)
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.

Is this approach correct ?

Kudos [?]: 73 [0], given: 49

Intern
Status: Single
Joined: 12 Oct 2014
Posts: 7

Kudos [?]: 8 [0], given: 53

Location: India
Concentration: Strategy, Operations
GMAT Date: 11-11-2014
GPA: 3.1
WE: Information Technology (Computer Software)
Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]

### Show Tags

14 Nov 2014, 06:26
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

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.

Kudos [?]: 8 [0], given: 53

Intern
Joined: 14 Apr 2015
Posts: 18

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

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

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.

Hello
I think 2^32 = (2^2)^16 =(4)^16 =(3+1)^16

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

Manager
Status: One more try
Joined: 01 Feb 2015
Posts: 52

Kudos [?]: 20 [0], given: 154

Location: India
Concentration: General Management, Economics
WE: Corporate Finance (Commercial Banking)
Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]

### Show Tags

13 Dec 2015, 07:36
32 can be written as 4k+4
k being multiple of 7
32^4k+4^4k+4
=32^4^4
=32^64
=32^9k+1
=32
32 div by 7 gives rem 4
_________________

Believe you can and you are halfway there-Theodore Roosevelt

Kudos [?]: 20 [0], given: 154

Intern
Joined: 29 May 2015
Posts: 6

Kudos [?]: 2 [0], given: 161

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

### Show Tags

28 Jan 2016, 15:30
1
This post was
BOOKMARKED
VeritasPrepKarishma

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!!

Kudos [?]: 2 [0], given: 161

Intern
Joined: 05 Jun 2015
Posts: 25

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

Location: Viet Nam
GMAT 1: 740 Q49 V41
GPA: 3.66
Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]

### Show Tags

11 Feb 2016, 09:00
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!

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

Intern
Joined: 18 Mar 2015
Posts: 9

Kudos [?]: 1 [0], given: 65

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

### Show Tags

11 Feb 2016, 13:43
32^32^32
I took first 32^32 . 32/4 cyclicity i took . Then i took 32^32 again . Similarly i got the final figure as 32/7 . The remainder is 4.

Kudos [?]: 1 [0], given: 65

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7611

Kudos [?]: 16905 [0], given: 230

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

### Show Tags

11 Feb 2016, 22:31
truongynhi wrote:
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!

Yes, if given a choice, one should prefer addition - not only is it easier, it's less prone to error. Here, (28+4) and (35 - 3) were equivalent so I used (28+4). Obviously, in a case such as (35 - 1), one would use subtraction.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

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

Veritas Prep Reviews

Kudos [?]: 16905 [0], given: 230

Intern
Joined: 23 Sep 2015
Posts: 47

Kudos [?]: 3 [0], given: 99

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

### Show Tags

16 Feb 2016, 07:25
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?????

Kudos [?]: 3 [0], given: 99

Math Forum Moderator
Joined: 02 Aug 2009
Posts: 4909

Kudos [?]: 5220 [2], given: 112

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

### Show Tags

16 Feb 2016, 07:57
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

Kudos [?]: 5220 [2], given: 112

Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2684

Kudos [?]: 1682 [0], given: 792

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]

### Show Tags

16 Mar 2016, 07:38
MeghaP wrote:

4^32^32 can also be solved by binomial theorem which would result in the remainder one.

2^2^32^32 = 2^3^16^32 =
8^16^32 = (7+1)^16^32
and hence the remainder will be one.

I am not sure where I am going wrong (text in red above is not correct). Please suggest.!!

You are not breaking down $$4^{{32}^{32}}$$ correctly.

It should rather be,

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

Now, follow what Bunuel has done in his solution at what-is-the-remainder-when-32-32-32-is-divided-by-100316.html#p774902

Hope this helps.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Kudos [?]: 1682 [0], given: 792

BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2222

Kudos [?]: 797 [0], given: 595

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

### Show Tags

17 Mar 2016, 00:11
Looked ...Re looked ...RE RE looked...
NO where to go
chetan2u you gotta help with this..
Regards
_________________

Give me a hell yeah ...!!!!!

Kudos [?]: 797 [0], given: 595

Math Forum Moderator
Joined: 02 Aug 2009
Posts: 4909

Kudos [?]: 5220 [1], given: 112

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

### Show Tags

17 Mar 2016, 02:07
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

Kudos [?]: 5220 [1], given: 112

Math Expert
Joined: 02 Sep 2009
Posts: 41652

Kudos [?]: 124262 [1], given: 12077

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

### Show Tags

20 Oct 2016, 23:28
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;
...
_________________

Kudos [?]: 124262 [1], given: 12077

Senior Manager
Joined: 13 Oct 2016
Posts: 367

Kudos [?]: 388 [0], given: 40

GPA: 3.98
What is the remainder when 32^32^32 is divided by 7? [#permalink]

### Show Tags

27 Nov 2016, 11:48
[quote="gurpreetsingh"]What is the remainder when $$32^{32^{32}}$$ is divided by 7?

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

$$32 = 4 (mod 7)$$

(4^32^32)/7 (I wasn't able to make it proper looking composite power. Sorry.)

$$7$$ is prime, $$GCF (4; 7) = 1$$. So $$4^6 = 1 (mod 7)$$

$$\frac{32^{32}}{6} = \frac{(2^5)^{32}}{6} = \frac{2^{160}}{6} = \frac{2^{159}}{3}$$

$$2 = -1 (mod 3)$$

$$\frac{(-1)^{159}}{3} = \frac{-1}{3} = -1 (mod 3) = 2 (mod 3)$$

Multiplying by cancelled factor $$2$$: = $$4 (mod 6)$$

$$\frac{1 * 4^4}{7} = \frac{2^8}{7}$$

$$2^3 = 1 (mod 7)$$

$$\frac{(2^3)^2*2^2}{7} = \frac{1*4}{7}$$

Our remainder is $$4$$.

Kudos [?]: 388 [0], given: 40

What is the remainder when 32^32^32 is divided by 7?   [#permalink] 27 Nov 2016, 11:48

Go to page   Previous    1   2   3   [ 55 posts ]

Similar topics Replies Last post
Similar
Topics:
6 What is the remainder when 7^9 - 4 is divided by 7? 5 14 Apr 2017, 08:35
1 If n divided by 7 has a remainder of 2, what is the remainder when 3 8 03 Jun 2017, 10:28
10 What is the remainder when 5^68 is divided by 7? 8 15 May 2017, 10:21
3 What is the remainder when 7^442 is divided by 10? 5 24 Sep 2016, 22:14
39 What is the remainder when 333^222 is divided by 7? 19 02 Dec 2016, 03:32
Display posts from previous: Sort by