If a googol is 10^100 and a googolplex is 10^googol, what is the digit immediately preceding the last...

[ Editor: the puzzle was posted by Mike Fee <M.Fee@irl.cri.nz>, who
spelled the words "googal" and "googalplex"; Gareth Rees pointed out
the spelling, and two other replies guessed 5 and 3. ]

The answer to the puzzle is "zero".  In fact, not only is that
particular digit zero, but all the next 69897E95 digits of that
monstrous number are also zero.

Hint: If G is any integer greater than 1, show that 2^(10^G + G + 1)
and 2^(G+1) have the same last (G+1) digits in base 10.  A proof by
induction on G is not difficult.  In Mike Fee's puzzle, we use G =
10^100 = a googol.  Then it is easy to determine the required digits of
2^(G+1), since 2^(G+1) is so much smaller than 10^(G+1).

It is a known result that any specified finite string of digits will
occur in some (usually large) power of 2.  The proposer came up with an
interesting explicit formula for one such power when the digits
specified are all zeroes.

A slightly more general result: for positive integers A, B and N, when
do 2^A and 2^(A+B) have the same last N digits in base 10?  Answer:
A >= N, and B is divisible by 4*5^(N-1)

-- Barry Wolk <wolk@ccm.UManitoba.CA>  (October 1994)