Question

Two problems: What are the last 2 digits of 3^n and 3^(3^n), 3^((3^(3^n), etc. Thanks.

For 2^n, 2^(2^n), 2^((2^(2^n) they are quite simple: 04, 08, 12, 16, 24, 28, 32, 36, 44, 48, 52, 56, 64, 68, 72, 76, 84, 88, 92, 96 and 2^(2^n): 16, 56, 96, 36, 2^((2^(2^n): 36.

I can't seem to find this pattern with powers of 3, that is if there is such a pattern.

Answer 1

@ganeshie8 if you are here, please, give him your elegant method and result. :)

Answer 2

I have a problem understanding the question here. What is the value of n here ?

Answer 3

To me, n is any number. So, we have a set of 2 digit numbers as a result.

Answer 4

I want this sequence https://oeis.org/A206636 for base 3? If you answered my previous problem that would help.

Answer 5

@ganeshie8 hi dear my friend - please help me getting back my profile and presentlie i not can sending to somebody messages bc, the submit button was blocked

Answer 6

Oh sorry about that glitch. I'll message @Preetha

Answer 7

ty and please say to about submit blocked button too - ty. in advance

Answer 8

Can you tell me how 08 is in your sequence of 2^n ?

Answer 9

2^3 = 08; 2^23 = ...08, 2^43 = ...08,

Answer 10

Don't you just want the last two digits of powers of 2?

Answer 11

the last 2 digits of 3^n and 3^(3^n)

Answer 12

How does 2^3 fit in that pattern ?

Answer 13

2
2^2
2^(2^2)
....

Answer 14

I don't see 2^3

Answer 15

I am looking for a complete cycle for the last 2 digits of 3^n, the first period is 4, with 1, 3, 7, 9. I can't find the second period.

Answer 16

What has knuthh up arrow got anything to do with it ?

Answer 17

https://en.m.wikipedia.org/wiki/Knuth%27s_up-arrow_notation

Answer 18

n^^a = n^n^n...n with a-1 ^ marks, n^^1 = n, n^^2 = n^n, n^^3 = n^n^n, etc....

Answer 19

arent you mixing powers of 2 with tetration

Answer 20

I don't get why you're considering 2^3
It is not a tetration

Answer 21

Nope. Just taking 2^^n mod 10^(n-1), except I want this with 3^^n mod 10^(n-1), and that is too hard to compute after n = 3.

Answer 22

Again, 2^3 doesnt beling in the sequence :

2
2^2
2^(2^2)
....

Answer 23

But I think I see what you're doing. Let me think a bit...

Answer 24

For 3^^n you should get just two remainders :

27, 87

Answer 25

Which one is the last remainder that continues in this sequence: 7, ?, ?, ?,

Answer 26

Try proving below result :

3^^4 = 3^^3 mod 100

Answer 27

3^^4 is a few trillion digits.

Answer 28

Yes, but you don't really need to write out the number in order to prove above result

Answer 29

I can't think of any way to compute that.

Answer 30

Can you work 3^(3^3) mod 100 ?

Answer 31

Use this fact :
3^20 = 1 mod 100

Answer 32

Yes. 87. This would be the case for an number of the form 3^(20x+7)

Answer 33

Yes
Next work 3^^4 mod 100

Answer 34

3^(3^(3^3)) mod 100

Answer 35

The fact that the tens place is always even, so 3^n should have the form 20x+1,3,7,9?

So remainders are 01, 03, 07, 09, 21, 23, 27, 29, 41, 43, 47, 49, 61, 63, 67, 69, 81, 83, 87, 89 is my guess? Not sure. The next step I should have four remainders: 03, 27, 83, 87? Did I get this right?

Answer 36

Again why are you mixing powers of 3 and tetrations?

Answer 37

This is the first two steps: 27, 87. Then 87 is my last remainder. So my first few sequences start here: 7, 87, ?. To find the third term would be either 087, 187, 287, 387, 487, 587, 687, 787, 987?

Answer 38

You want to find last 3 digits also is it ?

Answer 39

Yes. But first how do I prove the tens place is always even? I just showed that is most likely.

Answer 40

It may be useful to know, that by basic number theory, \(3^{99} \equiv 1\mod100\) by Euler's Theorem.

Answer 41

3^100 = 1 (mod 100)?

Answer 42

By quadratic residues, every odd square should be congruent to 1 (mod 8)? 3^2 = 1 (mod 8) and 3^3 = 3 (mod 8). The last three digits that end in 3 or 7 should be congruent to 3 (mod 8)? That leaves my understanding that the next term is either 187, 387, 587, 787, or 987?

Answer 43

Ooops, sorry, \(3^{40} \equiv 1 \mod 100\) and not \(3^{99} \equiv 1\). Euler's Theorem: https://en.wikipedia.org/wiki/Euler%27s_theorem

Answer 44

http://m.wolframalpha.com/input/?i=Table%5B3%5En+mod+100%2C+%7Bn%2C+1%2C+19%7D%5D&x=0&y=0

Answer 45

http://m.wolframalpha.com/input/?i=Table%5B3%5E3%5En+mod+100%2C+%7Bn%2C+1%2C+10%7D%5D&x=0&y=0