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ganeshie8
unit's digit of \(\large 3^{3^{3^{3^{3^{3^{3}}}}}} \)
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
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
wolfram says 7 though http://www.wolframalpha.com/input/?i=3%5E3%5E3%5E3%5E3%5E3%5E3%5E3
3 and 7 will repeat..
3,9,7,1 so it repeats every fourth power
how did u conclude "1" @BSwan
``` 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
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 :)
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
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
why there is water in eyes anyway ? that remindes me with a novel memories of geisha , did u read it ?
There are 8 3s in wolfram... But answer is yet the same..
@myininaya we need to work it from top/exponent right, pattern might be tough to predict for the last digit i thinik
No I have not @BSwan
why the answer is same cuz 3=-1 mod 4 thus 3^(any odd power )=-1 mod 4
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...
oh you know what i was thinking about the first digit my bad
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
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"
units tens hundred thousands... yes sure, this is the correct terminologhy :)
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
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
Now I see :) but you seem to be starting from the bottom term first ? \[ \large \color{Red}{3}^{\color{red}{3}^{3^{3^{3^{3^{3}}}}}} \]
i see haha! i think it would be bit more hard if we had to work it from the top first
I think these type of problems aren't normally looked at the way I approached but with the whole mod thing
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
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 ?
what i meant is, working units digit of some number like : (3408504385094385)^3 is easy compared to 3^3^3^3^3^3
since it is a single exponent, you can work it in a snap
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
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
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
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
they're two different problems, and two different methods. your method and the expression are challenging myininaya's method and the expression are simpler
you need to work on ur vocablary a bit :P challenging is not a negative work
similar problem+interesting dicussions @ http://math.stackexchange.com/questions/43327/evaluate-the-last-digit-of-77777
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.
ur not a moron @myininaya ;)
only one moron could be exist ;)
i think in the arab world last digit is left most as they start writing from right side of the page xD
The units digit is normally the last digit to be written
I think well sometimes
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
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
I would like to know more about @bswan 's approach We haven't finish that way right?
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
I'll explain Bswan's method quick
it uses below fact \(\large 3^{4k}\) has last digit of \(\large 1\) \(k\ge 1\)
\[\large 3^{\color{Red}{3^{3^{3^{3^{3^{3}}}}}}} = 3^{\color{Red}{4k+3}} = 3^3*3^{\color{Red}{4k}} = 27*1\]
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
yeah and simply you could just say 3=-1 mod 4 :P
thats exactly how i interpreted the method also xD
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 will pretend you are stabbing my cat because he keeps getting up in the sink after I have told him no several times
i feel it may not be possible to touch the left most digit
no @myininaya i cant do that i love cats Ps :- i dont stab any animals as will :P i only found it funny :3