unit's digit of
\(\large 3^{3^{3^{3^{3^{3^{3}}}}}} \)

- ganeshie8

unit's digit of
\(\large 3^{3^{3^{3^{3^{3^{3}}}}}} \)

- schrodinger

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- anonymous

awww to many 3's

- anonymous

chooo many 3's :3

- ganeshie8

7 threes only :)

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## More answers

- anonymous

well hmm let me think of a battern instead of calculation

- anonymous

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

- anonymous

I think 3 will the last digit...

- anonymous

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

- ganeshie8

3, 9, 7, 1...

- anonymous

9 ??

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

:3

- ganeshie8

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

- anonymous

3 and 7 will repeat..

- anonymous

xD

- ganeshie8

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

- ganeshie8

how did u conclude "1" @BSwan

- ganeshie8

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

- ganeshie8

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

- anonymous

wait ahahahaha yeah yeah which gives 7 :3

- anonymous

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

- anonymous

i love 7 anyway its hard number

- ganeshie8

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 :)

- anonymous

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

- anonymous

hehehehe

- anonymous

it does water in eyes :D

- ganeshie8

i would have added more 3's proabably haha!

- myininaya

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

- anonymous

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

- anonymous

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

- ganeshie8

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

- anonymous

No I have not @BSwan

- anonymous

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

- ganeshie8

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...

- myininaya

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

- ganeshie8

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

- anonymous

left most digit ?

- myininaya

wait

- anonymous

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

- myininaya

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

- anonymous

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

- myininaya

i'm confused about those terms first and last

- anonymous

Last digit is one's digit..

- ganeshie8

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

- anonymous

hmm shure ?

- ganeshie8

units
tens
hundred
thousands...
yes sure, this is the correct terminologhy :)

- anonymous

lavosh >.<

- myininaya

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

- myininaya

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

- anonymous

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

- anonymous

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

- myininaya

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

- ganeshie8

Now I see :) but you seem to be starting from the bottom term first ?
\[ \large \color{Red}{3}^{\color{red}{3}^{3^{3^{3^{3^{3}}}}}} \]

- myininaya

I was

- myininaya

or I did whatever

- ganeshie8

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

- myininaya

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

- ganeshie8

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

- ganeshie8

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

- ganeshie8

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

- anonymous

challenging ? hmm

- anonymous

well , its not you who can juge :P

- ganeshie8

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

- anonymous

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

- anonymous

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

- ganeshie8

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

- ganeshie8

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

- anonymous

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

- ganeshie8

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

- myininaya

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

- myininaya

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

- anonymous

see ,it depands on u :P

- ganeshie8

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

- ganeshie8

they're two different problems, and two different methods.
your method and the expression are challenging
myininaya's method and the expression are simpler

- anonymous

|dw:1408307650669:dw|

- ganeshie8

you need to work on ur vocablary a bit :P
challenging is not a negative work

- ganeshie8

*word

- anonymous

:*

- ganeshie8

similar problem+interesting dicussions @
http://math.stackexchange.com/questions/43327/evaluate-the-last-digit-of-77777

- myininaya

I don't know why the term last digit seems weird to me
but it does.
@ganeshie8 It looks like that term is used other places not just here.
:p
I feel like a moron a little.

- anonymous

ur not a moron @myininaya ;)

- anonymous

only one moron could be exist ;)

- ganeshie8

i think in the arab world last digit is left most as they start writing from right side of the page xD

- myininaya

The units digit is normally the last digit to be written

- myininaya

I think well sometimes

- myininaya

I have a good way to think about it
When we say the numbers in words the last digit we say anything about in those words are the units digit

- anonymous

yeah ganesh that is a problem to understand :P
left / right
lol

- anonymous

i would love to work on last digit though
its exited me for some reason

- myininaya

I would like to know more about @bswan 's approach
We haven't finish that way right?

- anonymous

were done already :o

- ganeshie8

thats a good analogy to relate to. however unit's digit looks less controversial to me lol... because for computer science ppl and many others first/last digit makes no sense when they save numbers in queues/stacks or other data structures

- ganeshie8

I'll explain Bswan's method quick

- ganeshie8

it uses below fact
\(\large 3^{4k}\) has last digit of \(\large 1\)
\(k\ge 1\)

- ganeshie8

\[\large 3^{\color{Red}{3^{3^{3^{3^{3^{3}}}}}}} = 3^{\color{Red}{4k+3}} = 3^3*3^{\color{Red}{4k}} = 27*1\]

- myininaya

I get that
That is beautiful
4k would not work because 3^some power or even 3 is not even
4k+1 would not work either because there is no integer k such that 4k+1=3
4k+2 would not work for the same explanation as 4k
4k+3 would work because 4k+3=3 when k=0
and 4k+3=27 when k=6 and so on...
Bswan, this is definitely pretty

- anonymous

yeah and simply you could just say
3=-1 mod 4 :P

- ganeshie8

thats exactly how i interpreted the method also xD

- anonymous

matches

- ganeshie8

thank you both xD it is some good learning to me :3

- myininaya

now i have no clue how to figure out the first number (the left most)
i bet that would be a killer

- anonymous

well at the end of this topic , i thought of sharing this xD

##### 1 Attachment

- myininaya

I will pretend you are stabbing my cat because he keeps getting up in the sink after I have told him no several times

- ganeshie8

i feel it may not be possible to touch the left most digit

- anonymous

no @myininaya i cant do that i love cats
Ps :- i dont stab any animals as will :P
i only found it funny :3

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