Re: A86: Tough math problem... or is it?

To: assembly-86@lists.ticalc.org
Subject: Re: A86: Tough math problem... or is it?
From: KidFedX@aol.com
Date: Wed, 11 Nov 1998 22:39:27 EST
Delivered-To: assembly-86-outgoing@towerguard.unix.edu.sollentuna.se
Reply-To: assembly-86@lists.ticalc.org


You realize that 2879 is a prime number and that a counting number will not
divide into it evenly.  This means that the "cents" will continue to get more
and more divided.
    All you have to do is take 2879 and divide it by the first ratio.  say 2.
that gives you 1430.5, then all you have to do is continue dividing that by
the ratio you want -- so your next # is 3.  giving you 479.83333... I don't
know how your teacher (I assume this is for class) wants you to do round-offs
but continue in this pattern and you will find that the person with the lowest
fortune has approximately $0.071403769841.  The answer (forgetting about
round-off errors) is that.  OR 2879/40320 OR 2879/8! (that's factorial).  so
with fractions of cents you are now set.  I don't think we should be doing
your math HW for you.
--Joe 
 PS just give me credit in front of your class :)

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