Posers and Puzzles

Posers and Puzzles

  1. Sigulda, Latvia
    Joined
    30 Aug '06
    Moves
    4048
    16 Feb '07 18:45
    Find the last digit in result of 2^1996.

    Explanation how you did it would be nice. It's not that I don't know the answer, I would just like to see if there are other ways than I figured of solving this problem.
  2. Joined
    21 Sep '05
    Moves
    75006
    16 Feb '07 19:00
    Originally posted by kbaumen
    Find the last digit in result of 2^1996.

    Explanation how you did it would be nice. It's not that I don't know the answer, I would just like to see if there are other ways than I figured of solving this problem.
    Answer below:
    #

    #

    #

    #

    #

    #

    #

    #

    #

    #

    The first five powers of 2 are: 2, 4, 8, 16, 32. The final digits will therefore cycle through 2, 4, 8, 6 (in that order). When the index is a multiple of 4 the final digit will be 6, hence 2^1996 has final digit of 6.
  3. Joined
    20 Jan '07
    Moves
    1005
    16 Feb '07 19:23
    Originally posted by kbaumen
    Find the last digit in result of 2^1996.

    Explanation how you did it would be nice. It's not that I don't know the answer, I would just like to see if there are other ways than I figured of solving this problem.
    6?
  4. Sigulda, Latvia
    Joined
    30 Aug '06
    Moves
    4048
    16 Feb '07 19:57
    Correct.
  5. Joined
    20 Jan '07
    Moves
    1005
    16 Feb '07 20:48
    Originally posted by kbaumen
    Correct.
    It was an easy one.



    The G


    p.s. Check "Site Ideas" for my revolutionary pole. Please make sure you vote.
  6. Standard memberPBE6
    Bananarama
    False berry
    Joined
    14 Feb '04
    Moves
    28719
    16 Feb '07 20:50
    Originally posted by GinoJ
    p.s. Check "Site Ideas" for my revolutionary pole.
    Ummm, no...I don't think I want to see your revolutionary "pole".
  7. Standard memberadam warlock
    Baby Gauss
    Ceres
    Joined
    14 Oct '06
    Moves
    18375
    16 Feb '07 23:30
    Originally posted by PBE6
    Ummm, no...I don't think I want to see your revolutionary "pole".
    ahahahahahaahahahahahahha
    ahahahahahahahahahahahahah
    ahahahahahahahahahahahahah

    Man you're crazy! 😀
  8. Joined
    20 Jan '07
    Moves
    1005
    16 Feb '07 23:31
    Originally posted by PBE6
    Ummm, no...I don't think I want to see your revolutionary "pole".
    😀
  9. Joined
    15 Feb '07
    Moves
    667
    18 Feb '07 16:08
    I'll, do you 3 better..

    The last 4 digits are 4336.
  10. Sigulda, Latvia
    Joined
    30 Aug '06
    Moves
    4048
    18 Feb '07 17:08
    Wow, I'm curious how did you figure that? Or is it just a random guess with being sure only about the last digit? Or you have a very advanced calculator?
  11. Joined
    15 Feb '07
    Moves
    667
    18 Feb '07 18:39
    Modulo arithmetic. The last 4 digits go through a cycle of 500 "last 4 numbers", and thus all I have to do is find the 500th and move backwards.

    To find that out, I determine that 2^500 modulo 625 is 1, and thus find the four digits that are divisible by 16 and 1 more than a multiple of 625. 625 modulo 16 is 1, so I'm looking for 625*15+1 or 9376.

    Now if I divide this by 16, I get another number, 586. So the remainder of my answer when divided by 625 has to be 586 AND divisible by 16.

    586 modulo 16 is 10, which means I want 6*625 + 586 which is 4336.

    (Belated math geek alert btw)
  12. Joined
    19 Oct '03
    Moves
    69376
    18 Feb '07 19:01
    Originally posted by geepamoogle
    Modulo arithmetic. The last 4 digits go through a cycle of 500 "last 4 numbers", and thus all I have to do is find the 500th and move backwards.

    To find that out, I determine that 2^500 modulo 625 is 1, and thus find the four digits that are divisible by 16 and 1 more than a multiple of 625. 625 modulo 16 is 1, so I'm looking for 625*15+1 or 9376.
    ...[text shortened]... dulo 16 is 10, which means I want 6*625 + 586 which is 4336.

    (Belated math geek alert btw)
    is the cycle through 500 numbers the same for all powers...like 3^n ect, or is the specific to 2^n. How is '500' found.
  13. Joined
    15 Feb '07
    Moves
    667
    18 Feb '07 19:27
    It's the same for 3.

    The theory is fairly simple. If you start with modulo 5, there are only 4 possible answers for relatively prime numbers (numbers not divisible by 5 in this case). So if you're looking at powers, it will start repeating after 4 consecutive powers. (For two it would be 1,2,4,3,1,2,4,3, or make it 10 and you get 1,2,4,8,6,2,4,8,6)

    Now with 25 (5 squared), you'll find that there are only 20 relatively prime numbers to look at, which means only 20 possiblities before it repeats itself.. (5,10,15,and 20 are all divisble by 5).

    With 125 (5 cubed), you see 100 relatively prime modulos, which is 5 times what you had with 25, which is 5 times what you had with 5 itself, so that pattern becomes obvious.
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