==> pickover/pickover.11.s <== ------------------------- Subject: Re: Cliff Puzzle 11: The Leviathan Number (PARTIAL SPOILER) Newsgroups: rec.puzzles References: <1992Oct21.135208.119425@watson.ibm.com> In article <1992Oct21.135208.119425@watson.ibm.com>, Cliff Pickover writes: > The leviathan number is defined as (10**666)!, where the "!" indicates > factorial. > 1. What are the first 6 digits of the leviathan number? The simplest technique would be to use Stirling's formula to compute the mantissa, i.e. frac( log(n) ) = frac( log(2*pi)/2 + log(n)/2 n*(log(n)-log(e)) ). In our case n = 10^666, so this equals frac( log(2*pi)/2 + 333 + 10^666*(666-log(e)) ) = frac( log(2*pi)/2 + 10^666*(1-log(e)) ), so we'd basically need to know something like 10 digits to the right of the decimal point for log(2*pi)/2, and something like 700 digits for log(e) (which is easily doable). We then compute (1-log(e)), shift the digits 666 spaces to the left, and we're all set. > 2. Could modern supercomputers compute the leviathan, or will this > beyond the realm of humankind for the next century? The number of digits is more than 10^668, and this compares unfavorably to the number of particles in the universe. Furthermore, even if a googol digits could be output per second, you'd never make it before the end of the universe. So, I'd say it's beyond the realm of humanity, period. > 3. Even if we cannot compute the leviathan, how many other > characteristics of this number can we write down. As another puzzle, how many zeroes does it end with, and what are the last two non-zero digits? .qq &EXIT THIS FILE HAS BEEN RECEIVED FROM BITNET The file may be executable. Before removing this header you must understand what the code will do. You must also have the appropriate intellectual property agreements in place before receiving the code into IBM. If you have any questions, contact your manager. The contents of the file has been shifted right by one character. Filename=(none) Filetype=(none) RECFM=F LRECL=80 Records=21 The file received from the BITNET gateway begins below the next line. ------------------------------------------------------------------------ Date: Thu, 22 Oct 1992 07:12 EDT From: Subject: RE: googol! To: CLIFF@YKTVMV Original_To: Jnet%"CLIFF@YKTVMV" Hi, Cliff. The log10(e) comes from applying Stirling's approximation for the factorial: for large n, n! is approximately sqrt(2*pi*n)*((n/e)^n). Substitute googol for n, take log10 of both sides, and recall the mantissa of the log10 gives the digits of the original number. In these days of fast symbolic packages allowing exact computation of large factorials (though presumably not so large as a googol), people forget Stirling's formula. Until a few years ago, this was the only way to find factorials (albeit, only approximately) for large numbers. Mike