2B. Determine all solutions to the equation

$$
3^{x}+3^{y}+3^{z}=21897
$$

such that $x, y, z \in \mathbb{N}$ and $x<y<z$.
Solution. We can write the equation in the form

$$
3^{x}\left(1+3^{y-x}+3^{z-x}\right)=27 \cdot 811
$$

Since $\left(3^{x}, 811\right)=1$ and $\quad\left(1+3^{y-x}+3^{z-x}, 27\right)=1$, from the last equation we get

$$
\left\{\begin{array}{l}
3^{x}=27 \\
1+3^{y-x}+3^{z-x}=811
\end{array}\right.
$$

From the first equation, we get $x=3$, and the second equation becomes

$$
\begin{aligned}
& 3^{y-3}+3^{z-3}=810 \\
& 3^{y-3}\left(1+3^{z-y}\right)=3^{4} \cdot 10
\end{aligned}
$$

Due to $\left(3^{y-3}, 10\right)=1$ and $\left(1+3^{y-z}, 3^{4}\right)=1$, we get

$$
\left\{\begin{array}{l}
3^{y-3}=3^{4} \\
1+3^{z-y}=10
\end{array}\right.
$$

From the first equation, we have $y-3=4$, i.e., $y=7$, and from the second, we get $3^{z-7}=3^{2}$.

Finally, the solution to the original equation is $x=3, y=7, z=9$.