What is the remainder for

(32^32^32)/7 ?

Will give the answers tomorrow.

Please post your answers with reasoning.

0.

32^32^32 can be written as
2^5*1024
= 2^5120

Now we find a pattern for powers of 2 divided by 7
2^0 = 0
2^1=0
2^2=0
2^3=1
2^4=2
2^5=0
2^6=1

Here the repetition starts. Every 3 nos. So 5120/3 gives a rmainder of 2. Since every 3rd power is 1. this should be 0

Ans. 0

Am not sure you read this correctly. ^ signifies raised to the power of. You have used multiplication(*) here or is there a super-duper simplification I just cant see

I feel that I need to explain the logic behind my above answer:

As every one knows, if we had multiplication in this problem it would be very easy to solve as vidyasagar pointed out.

(32*32*32)/7 = what is the remainder ?

To find the remainder, all you have to do is to neutralize the sevens and extract the remainders, and then to multiplay it, the result would then be:

4*4*4 = 64

the remainder is 1 = (7*9+1 = 64)

My logic is the same, just in bigger numbers (it's an easy way to find out if you master the basics ! people tend to get uneasy with big numbers)

32^32^32 = 32*32*32...... (n^32^32 times)

extratcing the remainders will give:

4*4*4*4*4....(n^32^32 times)

since 4^16 = 16^8

I can repeat this process to its basic elements and recover the remainder.

Aren't you suppose to use GP formula with exponents ?

2^5 does leave a remainder of 4.
But I'm still getting 4 as the remainde.
OA anyone?

I understand ! very nice , solving for AP where

AP = a + (n-1)d

a = first value = 1,2,3

d = difference = 3

will give you:

Group 1 (numbers with remainder 2) 5120 = 1 + (n-1)*3 ,n = 1,707,3333

Group 2 (numbers with remainder 4) 5120 = 2 + (n-1)*3, n = 1,707

Group 3 (numbers with remainder 1) 5120 = 3 + (n-1)*3, n = 1,706.666

so you claim that 2^5120 belongs to group 2 , hence with a remainder of 4,

this is a very clever thinking, but I think it's wrong !!

first of - your whole solution is based on the assumption that:

32^32^32 = 2^5120

I don't think this is correct ! can you please explain that logic ?

I took the liberty to summarize the above posts:

the remainder is clearly 4.

(32^32^32)/7 = what is the remainder ?

Method one

32^32^32 = 32^(32*32) = 2^(5*32*32) = 2^5120

2 power 1 - remainder 2
2 power 2 - remainder 4
2 power 3 - remainder 1
2 power 4 - remainder 2
2 power 5 - remainder 4
2 power 6 - remainder 1
and so on...

The progressions would be:

1,4,7...... is 5120 in here ? If so remainder 2
2,5,8...... is 5120 in here ? If so remainder 4
3,6,9...... is 5120 in here ? If so remainder 1

using Arithmetic Progression formula:

AP = a + (n-1)d

a = first value = 1,2,3

d = difference = 3

will give you:

Group 1 (numbers with remainder 2) 5120 = 1 + (n-1)*3 ,n = 1,707.3333
Group 2 (numbers with remainder 4) 5120 = 2 + (n-1)*3, n = 1,707
Group 3 (numbers with remainder 1) 5120 = 3 + (n-1)*3, n = 1,706.666

2^5120 is parst of group 2, hence the remainder is 4.

Method two

32^32^32

32^1024 where (32 is 7*4+4) and we need only the remainder:

4^1024 = 16^512 where (16 is 7*2+2) and we need only the remainder:

2^512 = 4^256 = 16^128 where (16 is 7*2+2) and we need only the remainder:

2^128 = 4^64 = 16^32 where (16 is 7*2+2) and we need only the remainder:

2^32 = 4^16 = 16^8 where (16 is 7*2+2) and we need only the remainder:

2^8 = 4^4 = 16^2 where (16 is 7*2+2) and we need only the remainder:

2^2 = 4

the remainder is 4.

thanks all (AdrianG, veekayem, Dek, vidyasagar, ggarr, waldeck55) for the fruitful discussion !!

we are still waiting for the OA !

Is this true in any case?

I think that the question ask for the remainder when (32^32^32)/7 not when (32*32*32)/7.