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ganeshie8

  • 9 months ago

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

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  1. BSwan
    • 9 months ago
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    awww to many 3's

  2. BSwan
    • 9 months ago
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    chooo many 3's :3

  3. ganeshie8
    • 9 months ago
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    7 threes only :)

  4. BSwan
    • 9 months ago
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    well hmm let me think of a battern instead of calculation

  5. waterineyes
    • 9 months ago
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    3, 7, 1, 3, 7, 1, 3, 7, 1 this will be the last digits for every corresponding powers..

  6. waterineyes
    • 9 months ago
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    I think 3 will the last digit...

  7. BSwan
    • 9 months ago
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    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
    • 9 months ago
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    3, 9, 7, 1...

  9. waterineyes
    • 9 months ago
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    9 ??

  10. BSwan
    • 9 months ago
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    so next step seeying 3^3^3^3^3^3 mod 4

  11. waterineyes
    • 9 months ago
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    Wait, let me think more, I am wrong..

  12. BSwan
    • 9 months ago
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    hm 3=-1 mod 4 thus 3^(odd ) =-1 mod 4 therfore checking for 4k+3 which gives 1 yeahhhh

  13. BSwan
    • 9 months ago
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    :3

  14. ganeshie8
    • 9 months ago
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    wolfram says 7 though http://www.wolframalpha.com/input/?i=3%5E3%5E3%5E3%5E3%5E3%5E3%5E3

  15. waterineyes
    • 9 months ago
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    3 and 7 will repeat..

  16. BSwan
    • 9 months ago
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    xD

  17. ganeshie8
    • 9 months ago
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    3,9,7,1 so it repeats every fourth power

  18. ganeshie8
    • 9 months ago
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    how did u conclude "1" @BSwan

  19. ganeshie8
    • 9 months ago
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    ``` 3=-1 mod 4 thus 3^(odd ) =-1 mod 4 therfore checking for 4k+3 ``` makes sense so far

  20. ganeshie8
    • 9 months ago
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    you get 3^(4k+3) finally, right ?

  21. BSwan
    • 9 months ago
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    wait ahahahaha yeah yeah which gives 7 :3

  22. BSwan
    • 9 months ago
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    xD i only counted 0,1,2,3 insted of 1,2,3,0 xD from battern

  23. BSwan
    • 9 months ago
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    i love 7 anyway its hard number

  24. ganeshie8
    • 9 months ago
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    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
    • 9 months ago
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    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
    • 9 months ago
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    hehehehe

  27. BSwan
    • 9 months ago
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    it does water in eyes :D

  28. ganeshie8
    • 9 months ago
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    i would have added more 3's proabably haha!

  29. myininaya
    • 9 months ago
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    I'm seeing a 7,3,7,3,... type of pattern

  30. BSwan
    • 9 months ago
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    why there is water in eyes anyway ? that remindes me with a novel memories of geisha , did u read it ?

  31. waterineyes
    • 9 months ago
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    There are 8 3s in wolfram... But answer is yet the same..

  32. ganeshie8
    • 9 months ago
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    @myininaya we need to work it from top/exponent right, pattern might be tough to predict for the last digit i thinik

  33. waterineyes
    • 9 months ago
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    No I have not @BSwan

  34. BSwan
    • 9 months ago
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    why the answer is same cuz 3=-1 mod 4 thus 3^(any odd power )=-1 mod 4

  35. ganeshie8
    • 9 months ago
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    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
    • 9 months ago
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    oh you know what i was thinking about the first digit my bad

  37. ganeshie8
    • 9 months ago
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    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
    • 9 months ago
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    left most digit ?

  39. myininaya
    • 9 months ago
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    wait

  40. BSwan
    • 9 months ago
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    hmm well find a battern of (3*10)^ something hehe cool lets think of it

  41. myininaya
    • 9 months ago
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    isn't the first digit the ones digit and the last digit the left-most digit

  42. waterineyes
    • 9 months ago
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    I am no good at MOD.. we were not taught this topic in mathematics anywhere in Schooling.. :)

  43. myininaya
    • 9 months ago
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    i'm confused about those terms first and last

  44. waterineyes
    • 9 months ago
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    Last digit is one's digit..

  45. ganeshie8
    • 9 months ago
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    Oh sorry, i should have used "unit's digit"

  46. BSwan
    • 9 months ago
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    hmm shure ?

  47. ganeshie8
    • 9 months ago
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    units tens hundred thousands... yes sure, this is the correct terminologhy :)

  48. BSwan
    • 9 months ago
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    lavosh >.<

  49. myininaya
    • 9 months ago
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    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
    • 9 months ago
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    I hope you guys know those ... just mean blah blah

  51. BSwan
    • 9 months ago
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    hmm ok -.- makes me feel we did nothing up there

  52. BSwan
    • 9 months ago
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    i thought ur gonna comment something fantacy like last digit >.< lavosh girl

  53. myininaya
    • 9 months ago
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    No no I wasn't trying to imply you guys were doing nothing

  54. ganeshie8
    • 9 months ago
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    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
    • 9 months ago
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    I was

  56. myininaya
    • 9 months ago
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    or I did whatever

  57. ganeshie8
    • 9 months ago
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    i see haha! i think it would be bit more hard if we had to work it from the top first

  58. myininaya
    • 9 months ago
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    I think these type of problems aren't normally looked at the way I approached but with the whole mod thing

  59. ganeshie8
    • 9 months ago
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    because 3^(3^3) gives 3^27 ****Bswan method**** (3^3)^3 gives 27^3 ****your method****

  60. ganeshie8
    • 9 months ago
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    your method is easy to work because we can take mod 10 readily for the bottom 27

  61. ganeshie8
    • 9 months ago
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    Bswan's method is bit more challenging as we don't know what happens in the exponent

  62. BSwan
    • 9 months ago
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    challenging ? hmm

  63. BSwan
    • 9 months ago
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    well , its not you who can juge :P

  64. ganeshie8
    • 9 months ago
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    i mean, compared to (27)^3, finding the unit's digit of 3^27 is challenging

  65. BSwan
    • 9 months ago
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    yeah right using simple theory is much chalenging than calculating 27^3 lol redicules jugment

  66. BSwan
    • 9 months ago
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    what if u have already base of 27^27^27^27 ?

  67. ganeshie8
    • 9 months ago
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    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
    • 9 months ago
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    since it is a single exponent, you can work it in a snap

  69. BSwan
    • 9 months ago
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    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
    • 9 months ago
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    however we cannot start from base for working the unit's digit of 3^(3^(3^(3^(3^(3^3)))))

  71. myininaya
    • 9 months ago
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    I don't think he was saying one way was better than the other

  72. myininaya
    • 9 months ago
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    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
    • 9 months ago
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    see ,it depands on u :P

  74. ganeshie8
    • 9 months ago
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    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
    • 9 months ago
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    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
    • 9 months ago
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    |dw:1408307650669:dw|

  77. ganeshie8
    • 9 months ago
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    you need to work on ur vocablary a bit :P challenging is not a negative work

  78. ganeshie8
    • 9 months ago
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    *word

  79. BSwan
    • 9 months ago
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    :*

  80. ganeshie8
    • 9 months ago
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    similar problem+interesting dicussions @ http://math.stackexchange.com/questions/43327/evaluate-the-last-digit-of-77777

  81. myininaya
    • 9 months ago
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    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
    • 9 months ago
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    ur not a moron @myininaya ;)

  83. BSwan
    • 9 months ago
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    only one moron could be exist ;)

  84. ganeshie8
    • 9 months ago
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    i think in the arab world last digit is left most as they start writing from right side of the page xD

  85. myininaya
    • 9 months ago
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    The units digit is normally the last digit to be written

  86. myininaya
    • 9 months ago
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    I think well sometimes

  87. myininaya
    • 9 months ago
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    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
    • 9 months ago
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    yeah ganesh that is a problem to understand :P left / right lol

  89. BSwan
    • 9 months ago
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    i would love to work on last digit though its exited me for some reason

  90. myininaya
    • 9 months ago
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    I would like to know more about @bswan 's approach We haven't finish that way right?

  91. BSwan
    • 9 months ago
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    were done already :o

  92. ganeshie8
    • 9 months ago
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    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
    • 9 months ago
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    I'll explain Bswan's method quick

  94. ganeshie8
    • 9 months ago
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    it uses below fact \(\large 3^{4k}\) has last digit of \(\large 1\) \(k\ge 1\)

  95. ganeshie8
    • 9 months ago
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    \[\large 3^{\color{Red}{3^{3^{3^{3^{3^{3}}}}}}} = 3^{\color{Red}{4k+3}} = 3^3*3^{\color{Red}{4k}} = 27*1\]

  96. myininaya
    • 9 months ago
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    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
    • 9 months ago
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    yeah and simply you could just say 3=-1 mod 4 :P

  98. ganeshie8
    • 9 months ago
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    thats exactly how i interpreted the method also xD

  99. BSwan
    • 9 months ago
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    matches

  100. ganeshie8
    • 9 months ago
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    thank you both xD it is some good learning to me :3

  101. myininaya
    • 9 months ago
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    now i have no clue how to figure out the first number (the left most) i bet that would be a killer

  102. BSwan
    • 9 months ago
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    well at the end of this topic , i thought of sharing this xD

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  103. myininaya
    • 9 months ago
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    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
    • 9 months ago
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    i feel it may not be possible to touch the left most digit

  105. BSwan
    • 9 months ago
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    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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