==> 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
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------------------------------------------------------------------------
 Date:     Thu, 22 Oct 1992 07:12 EDT
 From:     <FRAMEM@UNION>
 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