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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]

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New post 03 Sep 2010, 08:43
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]

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New post 03 Sep 2010, 10:19
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]

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New post 03 Sep 2010, 14:07
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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]

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New post 04 Sep 2010, 11:35
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]

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New post 04 Sep 2010, 12:31
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]

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New post 04 Sep 2010, 14:05
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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]

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New post 04 Sep 2010, 14:12
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]

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New post 04 Sep 2010, 15:03
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
Hence, answer is 3 here.

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Re: Tough remainder question [#permalink]

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New post 04 Sep 2010, 15:07
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]

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New post 04 Sep 2010, 15:39
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]

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New post 10 Oct 2010, 03:09
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]

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New post 17 Oct 2010, 16:25
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.
Could you please ellaborate more?
Thanks!
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Re: Tough remainder question [#permalink]

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New post 23 Mar 2012, 05:35
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]

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New post 19 Apr 2012, 10:51
gurpreetsingh wrote:
What is the remainder when \(32^{32^{32}}\) is divided by 7?

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

Please do not just post the answer, do explain as well.

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]

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New post 28 Apr 2012, 15:37
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]

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New post 14 Jul 2012, 09:27
Thanks so much for 2 excellent ways of solution of gurpreetsingh and Bunuel. I want to do some practice. :P

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)
B is the best answer.

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]

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New post 31 Jan 2013, 01:28
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: What is the remainder when 32^32^32 is divided by 7? [#permalink]

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New post 31 Aug 2013, 00:13
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

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

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New post 31 Jan 2014, 20:15
gurpreetsingh wrote:
What is the remainder when \(32^{32^{32}}\) is divided by 7?

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

Please do not just post the answer, do explain as well.

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.

Answer is B.

Took me 1:15 mins to solve.

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

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New post 05 Feb 2014, 01:09
gurpreetsingh wrote:
What is the remainder when \(32^{32^{32}}\) is divided by 7?

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

Please do not just post the answer, do explain as well.

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.
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Re: What is the remainder when 32^32^32 is divided by 7?   [#permalink] 05 Feb 2014, 01:09

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