Re: A86: Tough math problem... or is it?
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Re: A86: Tough math problem... or is it?
Im assuming that the F stands for the kid, and if there are only 7, then you
just need to find the N's....
In a message dated 11/11/98 10:02:43 PM Eastern Standard Time, QmH@aol.com
writes:
> I came across an interesting math problem. I couldn't think of anyway to
> approach it. If there's any of you math geniuses out there then try to
> give
> this problem a shot. The problem is as follows:
>
> Each of seven kids has a fortune consisting of a different number of
> dollars. The ratio of any kid's fortune to the fortune of any poorer kid
> will always be a counting number. The total value of the fortunes of all
> the kids is $2,879.00. What is the fortune of each kid?
>
> I can't think of any other way of doing this other than trial and error?
> Anybody know how to figure this one out?
>
> I came up with the general equation..
>
> F + F/(N1) + F/(N1*N2) + F/(N1*N2*N3) + F/(N1*N2*N3*N4) +
F/(N1*N2*N3*N4*N5)
> +
> F/(N1*N2*N3*N4*N5*N6) = 2879
>
> Try to solve for F when all the N's are counting numbers greater than 1 (
> since
> no two people can have the same fortune)..
>
> I couldn't do it.... my last resort is to do trial and error.. I'll make a
> program in VB or something that will try all the possible values of F and
> the
> N's using a bunch of FORs and then wait there for a while....
>
>
> ~ QmH
>