SOLUTION TO PROBLEM 8

The computation was possible because

$$\frac{1}{2}+\frac{1}{3}+\frac{1}{9}=\frac{17}{18}$$

hence the will stipulates what to be done with 17/18 of the property and leaves open the situation of 1/18 of it. Since 18is a multiple of 2, 3, and 9 and since 1/18 of 18 is exactly one, the judge could play his trick and recuperate his property.

Now how would one comply with the will and use calculus? Well, divide the remainder according to the will again, that is first calculate 1/2, 1/3, and 1/9 of 17 obtaining 17/2, 17/3, and 17/9 respectively, and a reminder of 17/18. Now this remainder should be split in three, the question being, how? Since the will says nothing more than 1/2 to the oldest 1/3 to the second oldest and 1/9 to the youngest one should proceed exactly that way and give $(1/2)\times (17/18)$ more to the first born, $(1/3)\times (17/18)$ to the second, and $(1/9)\times (17/18)$ to the youngest. By doing so one still has a remainder of 17/(18²). Continuing like that there will always be something left, so we should go like that for ever, that is continue infinitely many times. Assume we did just that what would happen? According to freshman calculus the first brother should get

$$\frac{1}{2}\left( 17+\frac{17}{18}+\frac{17}{18^2}+\frac{17}{18^3}+\cdots \frac{17}{18^n}+\cdots \right)=9.$$ (1)

Of course we used the geometric series formula saying that

$$\sum_{n=0}^\infty r^n=\frac{1}{1-r}$$

the series being convergent if and only if the ratio r is a number between -1 and 1.

A similar computation for the second brother would look as follows

$$\frac{1}{3}\left( 17+\frac{17}{18}+\frac{17}{18^2}+\frac{17}{18^3}+\cdots \frac{17}{18^n}+\cdots \right)=6.$$ (2)

The youngest should get then

$$\frac{1}{9}\left( 17+\frac{17}{18}+\frac{17}{18^2}+\frac{17}{18^3}+\cdots \frac{17}{18^n}+\cdots \right)=2.$$ (3)

So by applying the stipulations of the will ad infinitum one splits the heard exactly as the judge did, without tricks with extra camels added.

Questions and/or comments should be directed to Judy Downey or Griff Elder

    Back to the Problem of the Week Page
 
 

Last modified:   Thu Oct 18 14:24:47 CDT 2001