ganeshie8
  • ganeshie8
unit's digit of \(\large 3^{3^{3^{3^{3^{3^{3}}}}}} \)
Mathematics
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chestercat
  • chestercat
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anonymous
  • anonymous
awww to many 3's
anonymous
  • anonymous
chooo many 3's :3
ganeshie8
  • ganeshie8
7 threes only :)

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anonymous
  • anonymous
well hmm let me think of a battern instead of calculation
anonymous
  • anonymous
3, 7, 1, 3, 7, 1, 3, 7, 1 this will be the last digits for every corresponding powers..
anonymous
  • anonymous
I think 3 will the last digit...
anonymous
  • 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
  • ganeshie8
3, 9, 7, 1...
anonymous
  • anonymous
9 ??
anonymous
  • anonymous
so next step seeying 3^3^3^3^3^3 mod 4
anonymous
  • anonymous
Wait, let me think more, I am wrong..
anonymous
  • anonymous
hm 3=-1 mod 4 thus 3^(odd ) =-1 mod 4 therfore checking for 4k+3 which gives 1 yeahhhh
anonymous
  • anonymous
:3
ganeshie8
  • ganeshie8
wolfram says 7 though http://www.wolframalpha.com/input/?i=3%5E3%5E3%5E3%5E3%5E3%5E3%5E3
anonymous
  • anonymous
3 and 7 will repeat..
anonymous
  • anonymous
xD
ganeshie8
  • ganeshie8
3,9,7,1 so it repeats every fourth power
ganeshie8
  • ganeshie8
how did u conclude "1" @BSwan
ganeshie8
  • ganeshie8
``` 3=-1 mod 4 thus 3^(odd ) =-1 mod 4 therfore checking for 4k+3 ``` makes sense so far
ganeshie8
  • ganeshie8
you get 3^(4k+3) finally, right ?
anonymous
  • anonymous
wait ahahahaha yeah yeah which gives 7 :3
anonymous
  • anonymous
xD i only counted 0,1,2,3 insted of 1,2,3,0 xD from battern
anonymous
  • anonymous
i love 7 anyway its hard number
ganeshie8
  • 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
  • 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
  • anonymous
hehehehe
anonymous
  • anonymous
it does water in eyes :D
ganeshie8
  • ganeshie8
i would have added more 3's proabably haha!
myininaya
  • myininaya
I'm seeing a 7,3,7,3,... type of pattern
anonymous
  • anonymous
why there is water in eyes anyway ? that remindes me with a novel memories of geisha , did u read it ?
anonymous
  • anonymous
There are 8 3s in wolfram... But answer is yet the same..
ganeshie8
  • ganeshie8
@myininaya we need to work it from top/exponent right, pattern might be tough to predict for the last digit i thinik
anonymous
  • anonymous
No I have not @BSwan
anonymous
  • anonymous
why the answer is same cuz 3=-1 mod 4 thus 3^(any odd power )=-1 mod 4
ganeshie8
  • 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
  • myininaya
oh you know what i was thinking about the first digit my bad
ganeshie8
  • 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
  • anonymous
left most digit ?
myininaya
  • myininaya
wait
anonymous
  • anonymous
hmm well find a battern of (3*10)^ something hehe cool lets think of it
myininaya
  • myininaya
isn't the first digit the ones digit and the last digit the left-most digit
anonymous
  • anonymous
I am no good at MOD.. we were not taught this topic in mathematics anywhere in Schooling.. :)
myininaya
  • myininaya
i'm confused about those terms first and last
anonymous
  • anonymous
Last digit is one's digit..
ganeshie8
  • ganeshie8
Oh sorry, i should have used "unit's digit"
anonymous
  • anonymous
hmm shure ?
ganeshie8
  • ganeshie8
units tens hundred thousands... yes sure, this is the correct terminologhy :)
anonymous
  • anonymous
lavosh >.<
myininaya
  • 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
  • myininaya
I hope you guys know those ... just mean blah blah
anonymous
  • anonymous
hmm ok -.- makes me feel we did nothing up there
anonymous
  • anonymous
i thought ur gonna comment something fantacy like last digit >.< lavosh girl
myininaya
  • myininaya
No no I wasn't trying to imply you guys were doing nothing
ganeshie8
  • 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
  • myininaya
I was
myininaya
  • myininaya
or I did whatever
ganeshie8
  • ganeshie8
i see haha! i think it would be bit more hard if we had to work it from the top first
myininaya
  • myininaya
I think these type of problems aren't normally looked at the way I approached but with the whole mod thing
ganeshie8
  • ganeshie8
because 3^(3^3) gives 3^27 ****Bswan method**** (3^3)^3 gives 27^3 ****your method****
ganeshie8
  • ganeshie8
your method is easy to work because we can take mod 10 readily for the bottom 27
ganeshie8
  • ganeshie8
Bswan's method is bit more challenging as we don't know what happens in the exponent
anonymous
  • anonymous
challenging ? hmm
anonymous
  • anonymous
well , its not you who can juge :P
ganeshie8
  • ganeshie8
i mean, compared to (27)^3, finding the unit's digit of 3^27 is challenging
anonymous
  • anonymous
yeah right using simple theory is much chalenging than calculating 27^3 lol redicules jugment
anonymous
  • anonymous
what if u have already base of 27^27^27^27 ?
ganeshie8
  • 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
  • ganeshie8
since it is a single exponent, you can work it in a snap
anonymous
  • 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
  • ganeshie8
however we cannot start from base for working the unit's digit of 3^(3^(3^(3^(3^(3^3)))))
myininaya
  • myininaya
I don't think he was saying one way was better than the other
myininaya
  • 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
  • anonymous
see ,it depands on u :P
ganeshie8
  • 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
  • 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
  • anonymous
|dw:1408307650669:dw|
ganeshie8
  • ganeshie8
you need to work on ur vocablary a bit :P challenging is not a negative work
ganeshie8
  • ganeshie8
*word
anonymous
  • anonymous
:*
ganeshie8
  • ganeshie8
similar problem+interesting dicussions @ http://math.stackexchange.com/questions/43327/evaluate-the-last-digit-of-77777
myininaya
  • 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
  • anonymous
ur not a moron @myininaya ;)
anonymous
  • anonymous
only one moron could be exist ;)
ganeshie8
  • ganeshie8
i think in the arab world last digit is left most as they start writing from right side of the page xD
myininaya
  • myininaya
The units digit is normally the last digit to be written
myininaya
  • myininaya
I think well sometimes
myininaya
  • 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
  • anonymous
yeah ganesh that is a problem to understand :P left / right lol
anonymous
  • anonymous
i would love to work on last digit though its exited me for some reason
myininaya
  • myininaya
I would like to know more about @bswan 's approach We haven't finish that way right?
anonymous
  • anonymous
were done already :o
ganeshie8
  • 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
  • ganeshie8
I'll explain Bswan's method quick
ganeshie8
  • ganeshie8
it uses below fact \(\large 3^{4k}\) has last digit of \(\large 1\) \(k\ge 1\)
ganeshie8
  • 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
  • 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
  • anonymous
yeah and simply you could just say 3=-1 mod 4 :P
ganeshie8
  • ganeshie8
thats exactly how i interpreted the method also xD
anonymous
  • anonymous
matches
ganeshie8
  • ganeshie8
thank you both xD it is some good learning to me :3
myininaya
  • myininaya
now i have no clue how to figure out the first number (the left most) i bet that would be a killer
anonymous
  • anonymous
well at the end of this topic , i thought of sharing this xD
1 Attachment
myininaya
  • 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
  • ganeshie8
i feel it may not be possible to touch the left most digit
anonymous
  • 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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