Please turn on javascript in your browser to play chess.
Posers and Puzzles

Posers and Puzzles

  1. Standard member clandarkfire
    Grammar Nazi
    28 Apr '10 23:30
    What's the units digit of 13^(13^2010)?
  2. Subscriber joe shmo
    Strange Egg
    29 Apr '10 13:19
    Originally posted by clandarkfire
    What's the units digit of 13^(13^2010)?
    I can narrow it down to 3, or 7....having trouble getting it the rest of the way.
  3. 29 Apr '10 13:52
    Yes - it comes down to calculating 13^2010 mod 4. I'm sure there's some theorem or other that can help bring that down to something more manageable.
  4. Standard member Palynka
    Upward Spiral
    29 Apr '10 14:02 / 1 edit
    Originally posted by mtthw
    Yes - it comes down to calculating 13^2010 mod 4. I'm sure there's some theorem or other that can help bring that down to something more manageable.
    Why not 13^(13^(2010 mod 4) mod 4) = 13^(13^2 mod 4) = 13.

    That means 3 is the last digit...right? Maybe I'm missing something.
  5. Subscriber joe shmo
    Strange Egg
    29 Apr '10 14:06
    Originally posted by mtthw
    Yes - it comes down to calculating 13^2010 mod 4. I'm sure there's some theorem or other that can help bring that down to something more manageable.
    I dont know anything about modular arithmatic, so looks like im done. Just out of curiosity I know 13^2010 units digit ends in 9, will that help?
  6. Standard member clandarkfire
    Grammar Nazi
    30 Apr '10 22:28
    The answer is three

    Next question:

    How many digits does 13^(13^2010) have? An aproximate is obviously fine; its a big number.
  7. Standard member Agerg
    The 'edit'or
    01 May '10 02:13 / 5 edits
    Originally posted by clandarkfire
    The answer is three

    Next question:

    How many digits does 13^(13^2010) have? An aproximate is obviously fine; its a big number.
    screwed up...I'll try again
  8. Standard member Agerg
    The 'edit'or
    01 May '10 04:00 / 5 edits
    Originally posted by Agerg
    screwed up...I'll try again
    I might be wrong but expressing 13 as 10^log_{10}(13 )I'm thinking its about 10^29107
  9. Standard member clandarkfire
    Grammar Nazi
    01 May '10 04:54
    I honestly have no idea how to do this...could you explain your thinking a little?
  10. Standard member Agerg
    The 'edit'or
    01 May '10 12:36 / 5 edits
    Originally posted by clandarkfire
    I honestly have no idea how to do this...could you explain your thinking a little?
    Hmm...I think I was wrong, it was late and having malgebra troubles...anyway, lemme try again

    Since log_{10}(a^(b^c))=(b^c)*log_{10}(a) then
    c*log_{10}(b)+log_{10}(log_{10}(a))=log_{10}(log_{10}(a^(b^c)))

    Therefore
    log_{10}(log_{10}(13^(13^2010)))=2010*log_{10}(13)+log_{10}(log_{10}(13))~=2239

    giving that the correct answer should, I think have been roughly

    10^2239 digits
  11. Standard member clandarkfire
    Grammar Nazi
    01 May '10 16:49 / 1 edit
    Pretty much what I got...

    Use base-10 logarithms ("log" rather than "ln".)

    All two-digit integers have logs larger than or equal to 1, and less than 2.
    All three-digit integers have logs larger than or equal to 2, and less than 3.
    In general: all n-digit integers have logs larger than or equal to n-1, and less than n.

    Let x = 13^2010
    Then log x = log(13^2010)
    By log laws, the right-hand side becomes 2010*log(13), or roughly 2239.026138...
    So x has 2240 digits.

    You can write x as 10^2239.026138...
    Breaking the exponent into whole and fractional parts:
    x = 10^2239 * [10^0.026138136742...]
    which can be rearranged to
    = [1.062033...] * 10^2239
    in scientific notation.

    Now suppose y = 13^x
    Then log y = log(13^x)
    By log laws, log y = x*log(13)
    So log y = 1.18 * 10^2239


    1.18 * 10^2239 digits...Holy Christ.

    By comparison, there are roughly 8.0 * 10^80 atoms in the observable universe, (which is a spherical shape and has a radius of 46.5 billion light years or 2.73931666 × 10^23 miles.)
  12. Standard member Agerg
    The 'edit'or
    01 May '10 17:03
    Originally posted by clandarkfire
    Pretty much what I got...

    Use base-10 logarithms ("log" rather than "ln".)

    All two-digit integers have logs larger than or equal to 1, and less than 2.
    All three-digit integers have logs larger than or equal to 2, and less than 3.
    In general: all n-digit integers have logs larger than or equal to n-1, and less than n.

    Let x = 13^2010
    Then lo ...[text shortened]... a spherical shape and has a radius of 46.5 billion light years or 2.73931666 × 10^23 miles.)
    Heh...out of laziness and the convention followed by my lecturers and maths software I always take 'log' to mean the natural log (inspite of 'ln' so I specify base 10 logs by log_{10}

    But yeah, it's a bloody huge number!