# fibonacci puzzle problem

• Feb 3rd 2009, 11:35 AM
lauren2988
fibonacci puzzle problem
Three men possess a pile of money, their shares being ½, 1/3, 1/6. Each man takes some money from the pile until nothing is left. The first returns ½ of what he took, the second 1/3, and the third 1/6. When the total so returned is divided equally among the men, it is found that each posses what he is entitled. How much money was in the original pile and how much did each man take from the pile?
• Feb 3rd 2009, 04:41 PM
Soroban
Hello, lauren2988!

I did not come up with a unique solution . . .

Quote:

Three men possess a pile of money, their shares being $\tfrac{1}{2},\:\tfrac{1}{3},\:\tfrac{1}{6}$
Each man takes some money from the pile until nothing is left.
The first returns $\tfrac{1}{2}$ of what he took, the second $\tfrac{1}{3}$, and the third $\tfrac{1}{6}$.
When the total so returned is divided equally among the men,
it is found that each posseses what he is entitled.
How much money was in the original pile?
How much did each man take from the pile?

Let the three men be: . $A,B,C.$
. . And they took $a,b,c$ dollars, respectively.

$A$ returned half of his money.
. . He kept $\tfrac{a}{2}$ and returned $\tfrac{b}{2}$ dollars.

$B$ returned a third of his money.
. . He kept $\tfrac{2}{3}b$ and returned $\tfrac{b}{3}$ dollars.

$C$ returned a sixth of his money.
. . He kept $\tfrac{5}{6}c$ and returned $\tfrac{c}{6}$ dollars.

The total returned is: . $\frac{a}{2} + \frac{b}{3} + \frac{c}{6} \:=\:\frac{3a+2b+c}{6}$ dollars.

This was divided equally and returned to the men.
. . They each had: . $\frac{3a+2b+c}{18}$ dollars returned.

$A$ has: . $\frac{a}{2} + \frac{3a+2b+c}{18}$ . which is equal to his original share: . $\frac{a+b+c}{2}$

. . This simplifies to: . $3a - 7b - 8c \:=\:0$ .[1]

$B$ has: . $\frac{2}{3}b + \frac{3a+2b+c}{18}$ . which is equal to his original share: . $\frac{a+bc}{3}$

. . This simplifies to: . $3a - 8b + 5c \:=\:0$ .[2]

$C$ has: . $\frac{5}{6}c + \frac{3a+2b+c}{18}$ .which is equal to his original share: . $\frac{a+b+c}{6}$

. . This simplifies to: . $b - 13c \:=\:0$ .[3]

From [3], we have: . $b \,=\,13c$

Substitute into [1] and [2] and both of them become: . $a \,=\,33c$

Hence, there is no unique solution.

We have: . $\begin{Bmatrix}a &=& 33c \\ b &=& 13c \\ c&=&c\end{Bmatrix}$

On the right, replace $c$ with a parameter $t.$

. . and we have: . $\begin{Bmatrix}a &=& 33t \\ b &=& 13t \\ c &=& t\end{Bmatrix}$

The total amount of money is: . $33t + 13t + t \:=\:47t$ dollars.

Since the original amount was divisible by 6, $t$ must be a multiple of 6: . $t \,=\,6k$

And the parametric equations become: . $\begin{Bmatrix}a &=& 198k \\ b &=& 78k \\ c &=& 6k\end{Bmatrix}$

The total amount of money is: . $282k$ dollars ... for any positive integer $k.$