In each of the 16 circles in the figure, there is a student. A total of 3360 coins are distributed among the 16 students, such that students who are the same distance from the center receive the same number of coins. Five by five, from the outside to the inside, the circles are on the vertices of a regular pentagon, with the central circle equidistant from the vertices of the same pentagon.

![](https://cdn.mathpix.com/cropped/2024_05_01_09dbad5281c800d666d6g-26.jpg?height=505&width=521&top_left_y=673&top_left_x=649)

At a certain moment, each student gives all their coins, dividing them equally among each of their neighbors. After the exchange of coins, each student ended up with the same number of coins they had at the beginning. Determine the number of coins the student in the central circle originally had.
Solution

Denote each vertex with capital letters and the quantities each one passed by the respective lowercase letter. For example,

![](https://cdn.mathpix.com/cropped/2024_05_01_09dbad5281c800d666d6g-26.jpg?height=526&width=553&top_left_y=1757&top_left_x=633)

The 5 A's sent $4a$ and received $2a$ and $2b$, so $4a = 2a + 2b$ and consequently, $a = b$. The 5 B's sent $4b$ and received $2a$ and $2c$, thus $4b = 2a + 2c$. It follows that $b = c = a$. For the 5 C's, we have $3c = 2b + d = 2c + d$, hence $d = c = b = a$, all equal to $x$. The total amount distributed must match the total number of coins. Therefore,

$$
\begin{aligned}
4x \cdot 5 + 4x \cdot 5 + 3x \cdot 5 + 5x & = 3360 \\
60x & = 3360 \\
x & = 56
\end{aligned}
$$

Thus, the initial amount of the student in the central circle $D$ was $5x = 5 \cdot 56 = 280$ coins.