In article <··········@caleddon.dircon.co.uk> Simon Brooke <·····@rheged.dircon.co.uk> writes:
>·······@pegasus.cc.ucf.edu (Jason Soukeras) wrote:
>>Hey...I don't have access to lisp nor do I know how to program in it but
>>as far as I know it is the only language that will allow me to get the
>>mathematical answer i'm looking for.
>>Here is what I would like to know...
>>If I figure out what 1000! is
>>and divide it by a number that is 3 digits shorter... and then do the
>>same thing w/ another number that is 3 digits shorter but one more than
>>the first...will the portion of the answer to the left of the decimal
>>place be the same?
>
>>I realize I am asking a lot but would GREATLY appreciate it if someone
>>with 10 minutes or so on his/her hands would email me and let me know
>>or email me for clarification of the question I have if I haven't been clear.
Let m = 1000!, L = floor(log10(m)),
so 10^L <= m < 10^(L+1).
A number u is 3 digits shorter if int(log10(u)) = L-3
or 10^(L-3) <= u < 10^(L-2).
My interpretation of the question is this:
Is floor(m/u) = floor(m/(u+1)).
My answer: no!
Choose u so that it exactly divides m. For example,
choose u = m/1000. This is ok since m = 1000!.
Then m/u is an integer (in fact, 999!),
so that m/(u+1) < 999!.
If m is arbitrary and has no divisors up to 1000,
the question is a little harder.
--
Marty Cohen (······@nrtc.northrop.com) - Not the guy in Philly
This is my opinion and is probably not Northrop Grumman's!
Use this material of your own free will