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

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What is the remainder when 32^32^32 is divided by 7? [#permalink]  02 Sep 2010, 17:23
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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

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Last edited by gurpreetsingh on 04 Sep 2010, 10:43, edited 1 time in total.
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Re: Tough remainder question [#permalink]  02 Sep 2010, 18:16
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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.

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Last edited by ezhilkumarank on 02 Sep 2010, 18:32, edited 1 time in total.
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Re: Tough remainder question [#permalink]  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.
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Re: Tough remainder question [#permalink]  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.

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Re: Tough remainder question [#permalink]  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?
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Re: Tough remainder question [#permalink]  02 Sep 2010, 19:29
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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
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Re: Tough remainder question [#permalink]  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
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Re: Tough remainder question [#permalink]  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.
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Re: Tough remainder question [#permalink]  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
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Re: Tough remainder question [#permalink]  03 Sep 2010, 02:59
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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.
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Re: Tough remainder question [#permalink]  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.
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Re: Tough remainder question [#permalink]  03 Sep 2010, 05:33
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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.
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Re: Tough remainder question [#permalink]  03 Sep 2010, 06:51
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.
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Re: Tough remainder question [#permalink]  03 Sep 2010, 07:43
anshumishra : Its good if you know Euler theorem. The theorem is Fermat's little theorem that uses Euler theorem.

Guys you do not have to learn all these theorems for Gmat, if you know then its good if not then also you can solve this question using basics of remainders.Do not panic.

When 32^32 is divided by 6 the remainder is 4 not 2. Please check your solution.

32^32 when divided by 6 gives remainder same as when 2^32 is divided by 6

=> 2^32 mod 6 = $$(2^{5*{6}} )* (2^2)$$ mod 6 = 32^6 * 4 mod 6

= 2^6 *4 mod 6 = 2^5 * 2^3 mod 6 = 2*2 = 4
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Re: Tough remainder question [#permalink]  03 Sep 2010, 09:19
Thanks Gurpreet, I have edited my post where i missed to multiply by 4 instead of 2.
I have made "4" as bold 4.

Thanks
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Re: Tough remainder question [#permalink]  03 Sep 2010, 13:07
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mainhoon wrote:
How do you get 32x32= 6x +k?
anshumishra wrote:
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

Posted from my mobile device

You mean : How do you get 32^32= 6x +k?
Since, 32^6 mod 7 = 1
Hence 32^6x mod 7 = 1, that is why I am trying to express 32^32 in terms of 32^(6x+K), that way you have to just be concerned about calculating32^K mod 7

How it has been calculated is shown above.
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Re: Tough remainder question [#permalink]  04 Sep 2010, 10:26
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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.
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Re: Tough remainder question [#permalink]  04 Sep 2010, 10:35
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
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Re: Tough remainder question [#permalink]  04 Sep 2010, 10:44
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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.
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Re: Tough remainder question [#permalink]  04 Sep 2010, 11:31
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
Re: Tough remainder question   [#permalink] 04 Sep 2010, 11:31

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