Re: A86: Tough math problem... or is it?
[Prev][Next][Index][Thread]
Re: A86: Tough math problem... or is it?
KidFedX@aol.com writes:
> 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.
Actually, with this method applied to 7 kids, you get a total fortune of about
$2067.92. My (hopefully correct) answers are: $1450.83, $725.42, $362.71,
$181.35, $90.68, $45.34, and $22.67.
The equation to solve for F is a bit simpler than that in the original
message:
F + F/N1 + F/N2 + F/N3 + F/N4 + F/N5 + F/N6 = 2879
Since the ratio of each N to any with a higher number must be a counting
number, each N must be the LCM of the N's with lower numbers. The smallest
sequence you can get from this is: 2, 4, 8, 16, 32, 64. Solving for F with
these N's gives the above answers.
Be sure and tell your class that the nice people on Assembly-86 solved this
for you. ;)
--David