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awww to many 3's

chooo many 3's :3

7 threes only :)

well hmm let me think of a battern instead of calculation

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

I think 3 will the last digit...

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

3, 9, 7, 1...

9 ??

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

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

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

:3

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

3 and 7 will repeat..

xD

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

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

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

wait ahahahaha yeah yeah which gives 7 :3

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

i love 7 anyway its hard number

hehehehe

it does water in eyes :D

i would have added more 3's proabably haha!

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

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

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

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

left most digit ?

wait

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

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

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

i'm confused about those terms first and last

Last digit is one's digit..

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

hmm shure ?

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

lavosh >.<

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

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

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

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

I was

or I did whatever

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

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

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

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

challenging ? hmm

well , its not you who can juge :P

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

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

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

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

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

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

see ,it depands on u :P

|dw:1408307650669:dw|

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

*word

:*

ur not a moron @myininaya ;)

only one moron could be exist ;)

The units digit is normally the last digit to be written

I think well sometimes

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

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

were done already :o

I'll explain Bswan's method quick

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

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

thats exactly how i interpreted the method also xD

matches

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

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

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

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