Need expert help—fast? Use the Help Bell for personalized assistance getting answers to your important questions.

(32^32^32)/7

what would be the remainder.kindly help.

32^32^32 = 2 ^ 6 ^ 32 ^ 32 = 2 ^ (6 x 32 x 32) = 2 ^ 5120

I used my windows calculator to check if this trick is correct.

Now how about divided by 7?

google: arithmetic divided by 7

3rd link, "numbers and codes."

Page 13 and 14 of the PDF

I opened Excel and placed 2, 4, 8 and so on in one column and divided it by 7 in the next column.

See the REMAINDERS below.

2^1 --> 0.285714286

2^2 --> 0.571428571

2^3 --> 0.142857143

2^4 --> Now it repeats!

So now it becomes a question of dividing 5120 / 3 and what is the remainder?

5120 / 3 = 1706.6666

2 / 3 = 0.666666

So 2^5120 / 7 must have a remainder of 0.571428571!!

As additional proof for myself I checked with a small number on the calculator:

Remainder for 2^20 / 7 = 0.571428571

20 / 3 = 6.66666

I'm sure that for some math masters this is obvious but for me this fairly easy solution was a nice suprise.

this is how i worked it out

(28+4)^32^32/7

the last term in binomial expansion wud be 4^32^32.

now,

4^1/7 remainder is 4

4^2/7 remainder is 2

4^3/7 remainder is 1

we observe here cyclicity is 3

so (32^32)/3=(33-1)^32=3x+1

thus we have

4^3x+1/7

this will yeild remainder 1 with x assumed 1

is this approach correct?

In mod 7 arithmetic

32^32 = 4^32 = ((((4^2)^2)^2)^2)^2 ==> Squaring 5 times

The cycle I get is: 2 4 2 4 2 ==> 32^32 = 2 mod 7

Repeat the process once more:

(32^32)^32 = 2^32 = ((((2^2)^2)^2)^2)^2 ==> Squaring 5 times

The cycle I get is: 4 2 4 2 4 ==> 32^32^32 = 4 mod 7

32^n = 4^n

32^n = 4^n = 1 if and only if n is a multiple of 3

32^32^32 = (32^32)^32 = 32^(32^2) = 32^1024 <== There are no factors of 3 in that exponent

The answer is 4 as d-glitch said. I wrote it as a fraction as you can see. 4/7 = 0.57...

This step is correct.

4^1/7 remainder is 4 (line 1)

4^2/7 remainder is 2 (line 2)

4^3/7 remainder is 1 (line 3)

Your next step I don't understand.

1024/3 has remainder of 1, so you need to take line 1 from above, so 4^32^32 mod 7 = 4.

Did you read the PDF I told you about? It gives a lot of information specifically about this type of problem.

Also with examples and proof about mod n algebra addition and multiplication rules.

http://www.maths.ox.ac.uk/prospective-students/undergraduate/sutton/lecture1.pdf

For me this type of math is new so I took a simple try it and test it approach.

But from what I understand and try myself, d-glitch his story is perfect and also using the rules of this algebra.

My approuch is more figuring it out as I go.

=((2^5)^(32)^(32))

=(2^160)^(32)

=2^5120

You can notice that there is a pattern repeating on the last digit of the numbers when moded with 7:

2^1=2 = 2 mod 7

2^2=4 =4 mod 7

2^3=8 =1 mod 7

2^4=16 =2 mod 7

2^5=32 =4 mod 7

2^6=64 =1 mod 7

2^7=128 =2 mod 7

2^8=256 =4 mod 7

2^9=512 =1 mod 7

2^10=1024 =2 mod 7

You can see there is a pattern that repeats of 2, 4, 1, 2, 4, 1, 2, 4, 1 ....

i.e.

If a number 2^x = 2 mod 7

Then the number 2^(x+1) = 4 mod 7

If the number 2^x = 4 mod 7

Then the number 2^(x+1) = 1 mod 7

If a number 2^x = 1 mod 7

Then the number 2^(x+1) = 2 mod 7

which brings us back to the begining, i.e

If a number 2^x = 2 mod 7

Then the number 2^(x+1) = 4 mod 7

So if you want to work out what 2^n mod 7 is

Take the result of: n mod 3:

1) If the answer is 1 then 2^n mod 7 = 2

2) If the answer is 2 then 2^n mod 7 = 4

3) If the answer is 0 then 2^n mod 7 = 1

Now back to the result that 32^32^32 = 2^5120

5120 = 2 mod 3

So the answer from 3) is 4

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.

32^2 mod 7 = 4^2 mod 7 = 16 mod 7 = 2

32^4 mod 7 = 2^2 mod 7 = 4 mod 7 = 4

32^8 mod 7 = 4^2 mod 7 = 16 mod 7 = 2

32^16 mod 7 = 2^2 mod 7 = 4 mod 7 = 4

32^32 mod 7 = 4^2 mod 7 = 16 mod 7 = 2

(32^32)^2 mod 7 = 2^2 mod 7 = 4 mod 7 = 4

(32^32)^4 mod 7 = 4^2 mod 7 =16 mod 7 = 2

(32^32)^8 mod 7 = 2^2 mod 7 = 4 mod 7 = 4

(32^32)^16 mod 7 = 4^2 mod 7 =16 mod 7 = 2

(32^32)^32 mod 7 = 2^2 mod 7 = 4 mod 7 = 4