Let's find the integers $0 \leqq x_{1} \leqq x_{2} \leqq \ldots \leqq x_{9}$ that satisfy the following equality:

$$
3^{x_{1}}+3^{x_{2}}+\ldots+3^{x_{9}}=19683
$$
The powers of three that can be expressed in three terms are as follows:

| $\mathrm{k}$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $3^{k}$ | 1 | 3 | 9 | 27 | 81 | 243 | 729 | 2187 | 6561 |

$3^{9}$ cannot be used, as it equals 19,683. Let's categorize the possible solutions based on how many unknown values are equal to $x_{1}$. Denote this number by $M$, and $x_{1}$ itself by $m$. $M$ is clearly positive, and $M \leq 9$ because $m0$ cannot be. Therefore, $m<8$, and on the left side of (1) there are terms divisible by $3^{m+2}$. Since the right side is also divisible by $3^{m+2}$, $\left(\frac{M}{3}+N\right)$ must be divisible by 3.

If $M=6$, this implies $N=1$, and $S=3^{m+2}$ follows. The remaining two powers of 3 must sum up to $S$ to reach $3^{9}$, which is only possible if $S$ and these powers are all equal to $3^{8}$:

$$
x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=x_{6}=6, \quad x_{7}=7, \quad x_{8}=x_{9}=8
$$

If $M=3$, then $N$ can be 2 or 5, and $S$ can be $3^{m+2}$ or $2 \cdot 3^{m+2}$. In the second case, only one power of 3 remains, so it must also be equal to $3^{m+2}$, and $m$ is again 6:

$$
x_{1}=x_{2}=x_{3}=6, \quad x_{4}=x_{5}=x_{6}=x_{7}=x_{8}=7, \quad x_{9}=8
$$

In the first case, four of the unknowns are greater than $(m+1)$. Denote the number of these that are equal to $(m+2)$ by $K$. Since now $S+K \cdot 3^{m+2}=(K+1) 3^{m+2},(K+1)$ must be divisible by 3:

$$
x_{1}=x_{2}=x_{3}=5, \quad x_{4}=x_{5}=6, \quad x_{6}=x_{7}=7, \quad x_{8}=x_{9}=8
$$

Thus, the equation (1) has four solutions, labeled (2), (3), (4), and (5).

Note. If in a country the value of coins can only be a power of 3, and we have 9 coins each worth $3^{7}$, we can ask how many ways we can exchange our wealth while keeping it in 9 coins. Every time we exchange a coin in this country, we receive 3 coins of the next lower denomination, increasing our total by 2. We can reduce the number of coins by exchanging three of the same denomination for one of a higher value. According to this, we can obtain (4) from the solution labeled (2) by exchanging one coin and consolidating three others. From this, we can also obtain (3) and (5) by repeating this step, the only difference being that in (3) we exchange a $3^{7}$ coin, and in (5) a $3^{6}$ coin.