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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Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html

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

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

Was trying to use remainder theorem but just gave up :D
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Do you mean 32^{32^{32}} ? If yes, the answer is here
tough-remainder-question-100316-20.html

A new concept learnt. Thanks

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.

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.
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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)
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!

32^32^32 = 32^(2^160) = by remainder theorem, this is equal to 4^(2^160)

I will split the above term in term of 4^4( as 4^2 will give 16 and this in turn by remainder theorem will again give a 2)

Thus, 4^(2^160) = (4^4)^2^158 = (256)^2^158 = 4^(2^158)[by remainder theorem].

This can be repeated ,till in the end we will have only a 4. For eg,

(4^4)^2^156 = 4^(2^156) = and so on.

Thus remainder is 4.