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


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




On Wed, 11 Nov 1998 QmH@aol.com wrote:
> 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'm not sure I've understood the problem correctly, but if counting number
is the same thing as a positive integer (excuse me for being swedish
and not understand all english math terms...) the solution should be as
follows:

 Let f1, f2, .., f7 be the integer fortune of each kid, and let r1, r2,
 .., r6 be the ratio in such a way that f1*r1=f2, f2*r2=f3, ..., f6*r6=f7.
 Also, f1>0, r1,r2,..,r6> and finally f1+f2+...+f7=2879. (problem
 statement)

 Then  f1+f2+...+f7 = f1 + f1*r1 + f1*r1*r2 + ... + f1*r1*r2*r3*r4*r5*r6
                    = f1(1 + r1 + r1*r2 + ... + r1*r2*...*r6) = 2879

 Since 2879 is a prime, f1 must be 1. This simplyfies the equation to

   r1 + r1*r2 + ... + r1*r2*...*r6 = 2878 =>
   r1(1 + r2 + r2*r3 + ... + r2*r3*...*r6) = 2878 = 2*1439

 This makes r1=2 (1439 is a prime) and thus f1=2.

 This can be repeated through r2 to r5:
 
  r2(1+...)      = 1438 = 2*719 => r2=2 => f3=4
  r3(1+...)      =  718 = 2*359 => r3=2 => f4=8
  r4(1+r5+r5*r6) =  358 = 2*179 => r4=2 => f5=16
  r5(1+r6)       =  178 = 2*89  => r5=2 => f6=32
  
 And finally

  r6=88

 This gives f1=1, f2=2, f3=4, f4=8, f5=16, f6=32, f7=2816,
 and the sum of those just happens to equal 2879.

 (Note that this assumes the values searched for are integers;
  I think no other solution is possible due rounding problems)


If I misunderstood the problem; sorry!

-- 
Real name..: Jimmy Mårdell
E-mail.....: yarin@acc.umu.se
Homepage...: http://www.acc.umu.se/~yarin/



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