Question

unit's digit of

\(\large 3^{3^{3^{3^{3^{3^{3}}}}}} \)

Answer 1

awww to many 3's

Answer 2

chooo many 3's :3

Answer 3

7 threes only :)

Answer 4

well hmm let me think of a battern instead of calculation

Answer 5

3, 7, 1, 3, 7, 1, 3, 7, 1 this will be the last digits for every corresponding powers..

Answer 6

I think 3 will the last digit...

Answer 7

3=3 mod 10
3^2= 9 mod 10
3^3=7 mod 10
3^4=1 mod 10
3^5=3 mod 10
ohk fairr enough :3
4k+q

Answer 8

3, 9, 7, 1...

Answer 9

9 ??

Answer 10

so next step seeying 3^3^3^3^3^3 mod 4

Answer 11

Wait, let me think more, I am wrong..

Answer 12

hm
3=-1 mod 4
thus
3^(odd ) =-1 mod 4
therfore checking for 4k+3
which gives 1 yeahhhh

Answer 13

:3

Answer 14

wolfram says 7 though
http://www.wolframalpha.com/input/?i=3%5E3%5E3%5E3%5E3%5E3%5E3%5E3

Answer 15

3 and 7 will repeat..

Answer 16

xD

Answer 17

3,9,7,1
so it repeats every fourth power

Answer 18

how did u conclude "1" @BSwan

Answer 19

```
3=-1 mod 4
thus
3^(odd ) =-1 mod 4
therfore checking for 4k+3
```

makes sense so far

Answer 20

you get

3^(4k+3) finally, right ?

Answer 21

wait ahahahaha yeah yeah which gives 7 :3

Answer 22

xD
i only counted 0,1,2,3 insted of 1,2,3,0 xD
from battern

Answer 23

i love 7 anyway its hard number

Answer 24

Ahh that makes sense :) so basically you have worked it something like below :


3^3^3^3^3^3^3 = 3^(4k+3)
= 3^3*(3^4)^k
= 7(1)^k
= 7


nice :)

Answer 25

awww now i understand if its -1 why it keeps being -1 :o
all odd powers !
dint relize that on noon , im feel im much better now lol

Answer 26

hehehehe

Answer 27

it does water in eyes :D

Answer 28

i would have added more 3's proabably haha!

Answer 29

I'm seeing a 7,3,7,3,... type of pattern

Answer 30

why there is water in eyes anyway ?
that remindes me with a novel memories of geisha , did u read it ?

Answer 31

There are 8 3s in wolfram... But answer is yet the same..

Answer 32

@myininaya we need to work it from top/exponent right, pattern might be tough to predict for the last digit i thinik

Answer 33

No I have not @BSwan

Answer 34

why the answer is same cuz
3=-1 mod 4
thus
3^(any odd power )=-1 mod 4

Answer 35

3^3^3^3^3^3^3 = 3^(3^3^3^3^3)

its not same as (3^3)^3^3^3^3 or something... not so sure, need to check...

Answer 36

oh you know what i was thinking about the first digit my bad

Answer 37

Oh! never thought of left most digit before xD
last digit is easy to access using mod 10... but left most digit might be tough to work

Answer 38

left most digit ?

Answer 39

wait

Answer 40

hmm well find a battern of (3*10)^ something hehe
cool lets think of it

Answer 41

isn't the first digit the ones digit
and the last digit the left-most digit

Answer 42

I am no good at MOD.. we were not taught this topic in mathematics anywhere in Schooling.. :)

Answer 43

i'm confused about those terms first and last

Answer 44

Last digit is one's digit..

Answer 45

Oh sorry, i should have used "unit's digit"

Answer 46

hmm shure ?

Answer 47

units
tens
hundred
thousands...


yes sure, this is the correct terminologhy :)

Answer 48

lavosh >.<

Answer 49

well if it really was trying to figure out the units digit
then it would be easy right i think
\[3^3=..7\]
\[3^{3^3}=(..7)^3=\]
well since 7*7*7=49*7=...3
then we know
\[3^{3^3}=(..7)^3=...3\]


\[3^{3^{3^3}}=(...3)^3 \]
3*3*3=9*3=27
so we know

\[3^{3^{3^3}}=(...3)^3=...7 \]


this is how i got my 7,3,7,3,7,3... pattern

Answer 50

I hope you guys know those ... just mean blah blah

Answer 51

hmm ok -.-
makes me feel we did nothing up there

Answer 52

i thought ur gonna comment something fantacy like last digit >.<
lavosh girl

Answer 53

No no I wasn't trying to imply you guys were doing nothing

Answer 54

Now I see :) but you seem to be starting from the bottom term first ?

\[ \large \color{Red}{3}^{\color{red}{3}^{3^{3^{3^{3^{3}}}}}} \]

Answer 55

I was

Answer 56

or I did whatever

Answer 57

i see haha! i think it would be bit more hard if we had to work it from the top first

Answer 58

I think these type of problems aren't normally looked at the way I approached but with the whole mod thing

Answer 59

because


3^(3^3) gives 3^27 ****Bswan method****
(3^3)^3 gives 27^3 ****your method****

Answer 60

your method is easy to work because we can take mod 10 readily for the bottom 27

Answer 61

Bswan's method is bit more challenging as we don't know what happens in the exponent

Answer 62

challenging ? hmm

Answer 63

well , its not you who can juge :P

Answer 64

i mean, compared to (27)^3, finding the unit's digit of 3^27 is challenging

Answer 65

yeah right using simple theory is much chalenging than calculating 27^3 lol

redicules jugment

Answer 66

what if u have already base of 27^27^27^27
?

Answer 67

what i meant is, working units digit of some number like : (3408504385094385)^3
is easy compared to 3^3^3^3^3^3

Answer 68

since it is a single exponent, you can work it in a snap

Answer 69

ur opinion , its upto what you know and what u could memorize to solve something , i wont juge in anyway both looks cool to me , but i wont say something is better cuz bla blah blah

Answer 70

however we cannot start from base for working the unit's digit of 3^(3^(3^(3^(3^(3^3)))))

Answer 71

I don't think he was saying one way was better than the other

Answer 72

just because something is found to be more challenging doesn't mean it is a worse way
it means to me anyways that it is a way worth exploring

Answer 73

see ,it depands on u :P

Answer 74

exactly ! @BSwan i was saying the problems are different, so they both require different methods.

your method works for : 3^(3^(3^(3^(3^(3^3)))))
myininaya's method works for : ((((((3^3)^3)^3)^3)^3)^3

Answer 75

they're two different problems, and two different methods.


your method and the expression are challenging
myininaya's method and the expression are simpler