Three thieves $A, B$ and $C$ divided the proceeds of a robbery equally. On the first night, while $C$ was sleeping, $A$ and $B$ took half of what he had and divided it equally between themselves. On the second night, while $A$ was sleeping, $B$ and $C$ took half of what he had and divided it equally between themselves. On the third night, while $B$ was sleeping, $A$ and $C$ took half of what he had and divided it equally between themselves. The next morning, they parted ways forever. When $B$ counted his money, he realized he had $\mathrm{R} \$ 10,000.00$. Determine the initial amount that was stolen by the thieves.
Solution

Let $x$ be the amount of money each thief received in the initial division of the loot. On the first night, $x / 2$ of $C$'s money is divided into two portions of $1 / 2 \cdot x / 2 = x / 4$, one for $A$ and one for $B$. After this distribution, $A$ and $B$ each have $x + x / 4 = 5 x / 4$. The table below shows how much each one had on each night.

|  | $A$ | $B$ | $C$ |
| :--- | :---: | :---: | :---: |
| Initial | $x$ | $x$ | $x$ |
| Night 1 | $5 x / 4$ | $5 x / 4$ | $x / 2$ |
| Night 2 | $5 x / 8$ | $25 x / 16$ | $13 x / 16$ |
| Night 3 | $65 x / 64$ | $25 x / 32$ | $77 x / 64$ |

Since

$$
\begin{aligned}
\frac{25 x}{32} & =10,000 \\
x & =12,800
\end{aligned}
$$

Therefore, the initial amount was $3 \cdot 12,800 = \mathrm{R} \$ 38,400.00$.