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

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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
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Re: Tough remainder question [#permalink]  04 Sep 2010, 13:05
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.
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Re: Tough remainder question [#permalink]  04 Sep 2010, 13:12
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.
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Re: Tough remainder question [#permalink]  04 Sep 2010, 14:03
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
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Re: Tough remainder question [#permalink]  04 Sep 2010, 14:07
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
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Re: Tough remainder question [#permalink]  04 Sep 2010, 14:39
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.
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Re: Tough remainder question [#permalink]  10 Oct 2010, 02:09
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
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Re: Tough remainder question [#permalink]  17 Oct 2010, 15:25
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!
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Re: Tough remainder question [#permalink]  17 Oct 2010, 18:12
Was trying to use remainder theorem but just gave up :D
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Find the remainder when 32^32^32 is divided by 7? [#permalink]  28 Sep 2011, 00:47
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.
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Re: Find the remainder when 32^32^32 is divided by 7? [#permalink]  28 Sep 2011, 21:33
Do you mean 32^{32^{32}} ? If yes, the answer is here
tough-remainder-question-100316-20.html
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Re: Find the remainder when 32^32^32 is divided by 7? [#permalink]  29 Sep 2011, 01:14
nrgmat, thanks a lot. That's what I've been looking for.
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Re: Tough remainder question [#permalink]  19 Oct 2011, 20:35
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
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Re: Tough remainder question [#permalink]  19 Oct 2011, 22:25
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
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Re: Tough remainder question [#permalink]  23 Mar 2012, 04:35
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.
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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]  19 Apr 2012, 09:51
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.

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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]  28 Apr 2012, 14:37
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.
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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]  14 Jul 2012, 08:27
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!
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Re: Find the remainder when 32^32^32 is divided by 7? [#permalink]  31 Jan 2013, 00:28
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.

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Re: Tough remainder question [#permalink]  08 Jun 2013, 08:29
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, 08:29

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