This puzzle is in a vein similar to the powers
of ten and concerns factorials. The factorial function is useful
in many
areas of mathematics. It is defined by:
n! = n(n-1)(n-2)(n-3)...(3)(2)(1).
So, for example, 6! = (6)(5)(4)(3)(2)(1) = 720
and 10! =
(10)(9)(8)(7)(6)(5)(4)(3)(2)(1) = 3628800.
This week's puzzle is
to determine how many zeros occur at the end of 10000! .
Computing this number outright is difficult. It contains over
35,000 digits. If you are successful in determining how
many
zeros are at the end of 10000! you might wish to find a formula,
in terms of n, for the number of zeros
that occur at the end of
n! . If you tackle this, you may find the "integer part"
function useful. I(x) = the largest integer
less than or equal
to x. So I(2.3) = 2 and I(67.9) = 67 for example.