Puzzle
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Ponder This Challenge - September 2014 - 2^3^4^5 leading digit on base e

Ponder This Challenge:

Let b = (2^3^4^5) / (e^n), otherwise stated as "two to the power of (three to the power of (4 to the power of 5)) over e to the nth power", where n is an integer such that

1 < b < e.

Find b, with an accuracy of 10 decimal digits.

Update (9/11): There is a rather elegant way to solve this problem. We've added a star to solvers who found it.


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Solution

  • ln(2^3^4^5)=3^1024 * ln(2)

    We need to compute the fraction part of 3^1024 * ln(2).

    The Taylor series of ln(x) around 1 converges slowly, so to compute ln(2) one can compute ln(1/2) = -ln(2), which converges much faster; but to earn a star we use the following relation:

    ln(2) = ln(4/3)-ln(2/3)

    which gives us the series

    ln(2) = 2*sum_{i_0}^\infty 3^{-2i-1}\over(2i+1}

    3^1024 has 489 digits, so we need to compute ln(2) with ~500 digits accuracy, but using the above formula we can see that ~550 terms would suffice, since the contribution of the rest to 3^1024 ln(2) is much smaller than 10^-10. So we get that the fractional part is: 0.4261674320657868...

    The answer is the exponent of the above: 1.53137715531...

Solvers

  • Dan Dima (08/29/2014 03:57 PM EDT)
  • Dezhi Zhao (08/31/2014 04:42 AM EDT)
  • *Radu-Alexandru Todor (09/01/2014 07:27 AM EDT)
  • Daniel Bitin (08/31/2014 10:04 AM EDT)
  • Joaquim Neves Carrapa (08/31/2014 11:31 AM EDT)
  • Guangzhong Sun (09/01/2014 10:08 AM EDT)
  • avid Brink (09/01/2014 10:50 AM EDT)
  • Andreas Stiller (09/01/2014 12:51 PM EDT)
  • *Oleg Vlasii (09/01/2014 02:42 PM EDT)
  • Albert Stadler (09/01/2014 02:42 PM EDT)
  • Sumeet Katariya (09/01/2014 04:19 PM EDT)
  • Florian Fischer (EDT)
  • James Dow Allen (09/01/2014 08:33 PM EDT)
  • Joseph DeVincentis (09/01/2014 09:16 PM EDT)
  • John Snyder (09/01/2014 09:29 PM EDT)
  • Allen Sirolly (09/01/2014 09:41 PM EDT)
  • Vladimir Milovanovic (09/01/2014 11:58 PM EDT)
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  • Benjamin Phillabaum (09/02/2014 07:50 AM EDT)
  • Arthur Breitman (09/02/2014 07:55 AM EDT)
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  • Serge Batalov (09/02/2014 12:14 PM EDT)
  • David Dodson & Donald Dodson (09/02/2014 12:43 PM EDT)
  • Sean Egan (09/02/2014 12:49 PM EDT)
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  • *David Vainapel (09/27/2014 07:25 AM EDT)
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  • Seung Hwan (Sonny) An (09/28/2014 11:11 AM EDT)
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  • Carrie Oh (09/28/2014 04:44 PM EDT)
  • Kwonil Kobe Ko (09/28/2014 05:25 PM EDT)
  • Oskar Sigvardsson (09/29/2014 11:51 AM EDT)
  • Angel Mancebo (09/30/2014 03:14 PM EDT)
  • Konstantin Dzhigardzhyan (09/30/2014 03:57 PM EDT)
  • Karim Djaidja (10/01/2014 07:15 AM EDT)

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