3. On a chessboard, $8 \times 8$, there are 63 coins of 5 denari each and one coin of 10 denari, placed such that there is exactly one coin in each square. There are enough 5, 10, and 20 denari coins available. The following exchanges of three coins on the board with other three coins are possible:

$$
\begin{array}{ll}
(5,5,5) \leftrightarrow(10,10,10), & (5,5,10) \leftrightarrow(5,10,20) \leftrightarrow(20,20,10) \\
(5,10,10) \leftrightarrow(10,10,20), & (5,20,20) \leftrightarrow(5,5,20) \leftrightarrow(20,20,20)
\end{array}
$$

where the order does not matter.

Is it possible, after a finite number of exchanges on the board, to have 60 coins of 10 denari, 3 coins of 20 denari, and 1 coin of 5 denari?
Solution. Let's note that by replacing three coins on the board with three other coins, the sum of the values of the coins changes by 15 or 30. This means that the difference between the sums of the values of the coins on the board at the beginning and after several replacements is divisible by 15.

The sum of the values of the coins at the beginning is $63 \cdot 5 + 10 = 325$, while at the end of the desired arrangement it is $60 \cdot 10 + 3 \cdot 20 + 1 \cdot 5 = 665$. Their difference is 340, and this number is not divisible by 15.

Thus, the answer is no.