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Posers and Puzzles

Posers and Puzzles

  1. 12 Nov '04 07:58
    FIND THE LAST 10 DIGITS OF 9^(9^(9^9))
  2. Standard member TheMaster37
    Kupikupopo!
    12 Nov '04 10:45
    All I can tell is that it ends in a 9
  3. Standard member genius
    Wayward Soul
    12 Nov '04 11:08 / 4 edits
    i was bored so i attempted to find patterns with 9^whatnot...

    col 10=1,9,1,9 etc
    col 9=8,2,6,4,4,6,2,8,0,0 then repeats

    so if it's 9^an odd number, it'll end in 9.
    9^a even number, it'll end in 1.
    divide the (power-1) by 10, and if the remainders 1 then the second last digit of 9^ that number shall be a 8, if it's 2 it'll be a 2 etc (there are 10 numbers that repeat over)

    so, 9^(9^(9^9)).

    9^(9^9) is odd, so the last digit shall be 9. and 9^9 is 387420489, which is odd, so the second last digit of 9^(9^9) shall be 2. it'll end in 29. thus 9^(9^(9^9)) shall end in 89.

    i am bored. if you see a flaw, tell me. but from what i can see, that is correct...so, who's willing to find out the next 8 digits?

    coloum 8 goes (if we sub 1 for odd and 2 for even) 1122112222 and repeats...

    also-this is just using 9^1 to 25...
  4. Standard member Palynka
    Upward Spiral
    12 Nov '04 12:27
    Originally posted by cosmic voice
    FIND THE LAST 10 DIGITS OF 9^(9^(9^9))
    0123456789 ?
  5. Standard member squaccerman
    The 17th coming
    12 Nov '04 13:01
    "so if it's 9^an odd number, it'll end in 9.
    9^a even number, it'll end in 1."

    Surely if it ends in one it is not an even number.
  6. Standard member genius
    Wayward Soul
    12 Nov '04 13:37
    Originally posted by squaccerman
    "so if it's 9^an odd number, it'll end in 9.
    9^a even number, it'll end in 1."

    Surely if it ends in one it is not an even number.
    to the power an even number, e.g 9^124235432 shall end in 1, but 9^124235433 shall end in 9
  7. 12 Nov '04 15:27 / 1 edit
    Originally posted by Palynka
    0123456789 ?
    No that's not correct. Only the last two digits are correct. The process of finding the answer involves modular mathematics.
  8. 12 Nov '04 16:40
    Originally posted by cosmic voice
    FIND THE LAST 10 DIGITS OF 9^(9^(9^9))
    3111098189

    raise 9 to the power of a number of form kx10^n:
    - (n=0): The units digit cycles 1,9 and repeats
    - (n=1): The tens digit cycles every 10 (01,01,81,21,61,41,41,61,21,81)
    - (n>2): Subsequent digits cycle according to their digit.
    kx10^n (k=0,1,2,...,9; n=2,3,...,9) has last (n+1) digits of
    jx10^n+1 where j=0,4,8,2,6,0,4,8,2,6 for k = 0,1,2,3,4,5,6,7,8,9

    9^9 is 387420489
    9^9^9 has the same last 10 digits as 200000001.20000001.8000001.1.400001.1.601.21.9 which has the same last 10 digits as 2835673589 (obviously when you multiply you can just disgard any digits after the last 10).

    Repeat the process and you get the answer.
  9. 12 Nov '04 18:19 / 1 edit
    What are the last ten digits of (((((((9^9)^9)^9)^9)^9)^9)^9)^9?
  10. 13 Nov '04 00:01
    Originally posted by Glenton
    3111098189

    raise 9 to the power of a number of form kx10^n:
    - (n=0): The units digit cycles 1,9 and repeats
    - (n=1): The tens digit cycles every 10 (01,01,81,21,61,41,41,61,21,81)
    - (n>2): Subsequent digits cycle according to their digit.
    kx10^n (k=0,1,2,...,9; n=2,3,...,9) has last (n+1) digits of
    jx10^n+1 where j=0,4,8,2,6,0,4,8,2,6 for k = 0,1, ...[text shortened]... ou can just disgard any digits after the last 10).

    Repeat the process and you get the answer.
    Agree
  11. 16 Nov '04 14:17
    Originally posted by THUDandBLUNDER
    What are the last ten digits of (((((((9^9)^9)^9)^9)^9)^9)^9)^9?
    This number (given by you in your quoted post) is not very large number .....In fact it is miniscule when comparedto the number given in the puzzle.
  12. 16 Nov '04 20:40 / 2 edits
    4256697609


    6,261 digits in actual answer

  13. 17 Nov '04 17:35
    Originally posted by Gronk
    4256697609


    6,261 digits in actual answer

    NOPE...that is wide off the mark. you cant write all the digits of the number even if you have a scroll as wide as the earth's diameter and if
    each digit took just half a centimetre width.
  14. 21 Nov '04 12:16 / 1 edit
    Originally posted by cosmic voice
    NOPE...that is wide off the mark. you cant write all the digits of the number even if you have a scroll as wide as the earth's diameter and if
    each digit took just half a centimetre width.
    Whaoh......? REALLY ?
  15. Donation Acolyte
    Now With Added BA
    21 Nov '04 18:36 / 3 edits
    Originally posted by CoolPlayer
    Whaoh......? REALLY ?
    cosmic voice's number has more than (9^(9^9))/2 digits... and THAT has more than (9^9)/2 digits, so even writing down the number of digits of the number of digits of 9^(9^(9^9) in full would require you to write down millions of digits!

    THUDandBLUNDER's number is just 9^(9^8), which has relatively modest number of digits (fewer than 100 million).