$\frac{9}{10}$ is to be expressed as the sum of fractions, where the denominators of these fractions are powers of 6.
$$
\frac{9}{10}=\frac{3}{6}+\frac{2}{5}
$$

Thus,

$$
\frac{2}{5}=\frac{2}{6} \cdot \frac{6}{5}=\frac{2}{6} \cdot \frac{1}{\frac{5}{6}}=\frac{2}{6} \cdot \frac{1}{1-\frac{1}{6}}=\frac{2}{6}\left[1+\frac{1}{6}+\frac{1}{6}^{2}+\ldots\right]
$$

Therefore,

$$
\frac{9}{10}=\frac{3}{6}+\frac{2}{6}+\frac{2}{6^{2}}+\ldots=\frac{5}{6}+\frac{2}{6^{2}}+\frac{2}{6^{3}}+\ldots
$$

(Schwarcz Sándor, Beregszász.)

Note. It is easy to prove that $\frac{6}{5}$ can only be expressed in the form given above according to the problem.

Suppose it can also be expressed in the following form:

$$
\frac{6}{5}=a_{0}+\frac{a_{1}}{6}+\frac{a_{2}}{6^{2}}+\ldots+\frac{a_{n}}{6^{n}}+\ldots
$$

Then,

$$
a_{0}-1+\frac{a_{1}-1}{6}+\frac{a_{2}-1}{6^{2}}+\ldots+\frac{a_{n}-1}{6^{n}}+\ldots=0
$$

Since each fraction is non-negative, the equation can only hold if

$$
a_{0}=a_{1}=a_{2}=\ldots=a_{n}=\ldots=1
$$