Here's the math problem:
1*2*3*4*5*6*7*8*9...etc., etc. going up to 100.
How can I solve this problem without typing out all the numbers from 1
to 100?Is there some formula I can use?

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Request for Question Clarification by pafalafa-ga on 25 Oct 2004 07:10 PDT

Are you looking for the actual solution to 1*2*3*4*5*6*7*8*9...100, or
are you specifically in need of a formula for calculating the
factorial of 100?

I'd be happy to provide the solution, but as the comments below noted,
there is no simple shortcut (other than making use of calculator that
can compute factorial numbers).

Let us know what you're after.

pafalafa-ga

The following comments from http://answers.google.com/answers/threadview?id=33709
may be of some help.

4.6515988201944847664813222995241e+138

The product of the consecutive integers from 1 to 100 is called 100!,
but there is no specially effective shortcut to calculating it apart
from multiplying many times.

If we were asked instead to find the SUM of those same numbers, there
would be a nice way to simplify the work.  But no such help for the
PRODUCT.

regards, mathtalk-ga

As others have noted, the product of the consecutive integers from 1
to N is written as N!, and is "N factorial".  The value given by
evilben99 for 100! is *not* correct, though.  To four significant
figures, 100! is equal to 9.333 x 10^157

Mathtalk is correct that there is no "shortcut" for calculating the
*exact* value of N! other than multiplying out all the number;
however, there are approximations for calculating N! when N get large.
 The best known approximation is known as "Sirling's Approximation",
and is given by:
 
 N! ~= sqrt(2*pi*N)* N^N * exp(-N)

For 100!, this formula yields 9.325 x 10^157, which is an error of
about 0.085%.  As N gets larger, the approximation gets better.

See http://mathworld.wolfram.com/StirlingsApproximation.html
and
http://hyperphysics.phy-astr.gsu.edu/hbase/math/stirling.html

which provide some additional approximations that have somewhat smaller errors.