## ganeshie8 one year ago unit's digit of \(\large 3^{3^{3^{3^{3^{3^{3}}}}}} \)

1. BSwan

awww to many 3's

2. BSwan

chooo many 3's :3

3. ganeshie8

7 threes only :)

4. BSwan

well hmm let me think of a battern instead of calculation

5. waterineyes

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

6. waterineyes

I think 3 will the last digit...

7. BSwan

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

8. ganeshie8

3, 9, 7, 1...

9. waterineyes

9 ??

10. BSwan

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

11. waterineyes

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

12. BSwan

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

13. BSwan

:3

14. ganeshie8

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

15. waterineyes

3 and 7 will repeat..

16. BSwan

xD

17. ganeshie8

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

18. ganeshie8

how did u conclude "1" @BSwan

19. ganeshie8

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

20. ganeshie8

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

21. BSwan

wait ahahahaha yeah yeah which gives 7 :3

22. BSwan

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

23. BSwan

i love 7 anyway its hard number

24. 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 :)

25. BSwan

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

26. BSwan

hehehehe

27. BSwan

it does water in eyes :D

28. ganeshie8

i would have added more 3's proabably haha!

29. myininaya

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

30. BSwan

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

31. waterineyes

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

32. ganeshie8

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

33. waterineyes

No I have not @BSwan

34. BSwan

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

35. 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...

36. myininaya

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

37. 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

38. BSwan

left most digit ?

39. myininaya

wait

40. BSwan

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

41. myininaya

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

42. waterineyes

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

43. myininaya

i'm confused about those terms first and last

44. waterineyes

Last digit is one's digit..

45. ganeshie8

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

46. BSwan

hmm shure ?

47. ganeshie8

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

48. BSwan

lavosh >.<

49. 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

50. myininaya

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

51. BSwan

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

52. BSwan

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

53. myininaya

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

54. ganeshie8

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

55. myininaya

I was

56. myininaya

or I did whatever

57. ganeshie8

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

58. myininaya

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

59. ganeshie8

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

60. ganeshie8

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

61. ganeshie8

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

62. BSwan

challenging ? hmm

63. BSwan

well , its not you who can juge :P

64. ganeshie8

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

65. BSwan

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

66. BSwan

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

67. ganeshie8

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

68. ganeshie8

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

69. BSwan

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

70. ganeshie8

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

71. myininaya

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

72. 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

73. BSwan

see ,it depands on u :P

74. 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

75. 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

76. BSwan

|dw:1408307650669:dw|

77. ganeshie8

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

78. ganeshie8

*word

79. BSwan

:*

80. ganeshie8

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

81. 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.

82. BSwan

ur not a moron @myininaya ;)

83. BSwan

only one moron could be exist ;)

84. ganeshie8

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

85. myininaya

The units digit is normally the last digit to be written

86. myininaya

I think well sometimes

87. 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

88. BSwan

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

89. BSwan

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

90. myininaya

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

91. BSwan

were done already :o

92. 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

93. ganeshie8

I'll explain Bswan's method quick

94. ganeshie8

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

95. ganeshie8

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

96. 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

97. BSwan

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

98. ganeshie8

thats exactly how i interpreted the method also xD

99. BSwan

matches

100. ganeshie8

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

101. myininaya

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

102. BSwan

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

103. 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

104. ganeshie8

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

105. BSwan

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