28.33 The sum of nine fractions is 1, all of whose numerators are 1. Five of them are: $\frac{1}{3}, \frac{1}{7}, \frac{1}{9}, \frac{1}{11}, \frac{1}{33}$. The denominators of the other four fractions all end in 5. Then the denominators of these fractions are
(A) $5,15,35,105$.
(B) $5,25,35,135$.
(C) $5,15,45,385$.
(D) $5,25,105,385$.
（1st “Hope Cup” National Mathematics Invitational, 1990)
[Solution] Let the sum of the other four fractions be $S$, then
$$
S=1-\left(\frac{1}{3}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{33}\right),
$$
$\because$ The unit digits of the denominators of the other four fractions are all 5, so they are all multiples of 5, thus
$$
\begin{aligned}
S & =\frac{5}{5}\left[\left(1-\frac{1}{3}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{33}\right)\right] \\
& =\frac{1}{5}\left[5-\frac{5}{3}-\frac{5}{7}-\frac{5}{9}-\frac{5}{11}-\frac{5}{33}\right] .
\end{aligned}
$$

By estimation, $\frac{2}{12}\frac{5}{3}>5\left(1-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}-\frac{1}{11}-\frac{1}{33}\right)>\frac{10}{9}>1$,
Thus, if the terms in the parentheses are considered as the sum of four fractions, one of them must be 1, and the other three are less than 1, so
$$
\begin{aligned}
S & =\frac{1}{5}\left[1+\left(2-\frac{5}{3}\right)+\left(1-\frac{5}{9}\right)+\left(1-\frac{5}{7}-\frac{5}{11}-\frac{5}{33}\right)\right] \\
& =\frac{1}{5}\left[1+\frac{1}{3}+\frac{4}{9}+\left(\frac{2}{7}-\frac{5}{11}-\frac{5}{33}\right)\right],
\end{aligned}
$$

In the above expression, the numerator of $\frac{4}{9}$ is not 1, which is inappropriate. Rewrite it as $\frac{1}{9}+\frac{3}{9}$, then
$$
\begin{aligned}
S & =\frac{1}{5}\left[1+\frac{1}{3}+\frac{1}{9}+\left(\frac{2}{7}-\frac{5}{11}+\frac{3}{9}-\frac{5}{33}\right)\right] \\
& =\frac{1}{5}\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{77}\right) \\
& =\frac{1}{5}+\frac{1}{15}+\frac{1}{45}+\frac{1}{385} .
\end{aligned}
$$

Therefore, the answer is $(C)$.
Note 1: Fractions with a numerator of 1 are also known as "unit fractions" or "Egyptian fractions". The problem of expressing 1 as the sum of different unit fractions has the following conclusions:
$$
\begin{array}{l}
1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}, \quad \text { (fewest terms) } \\
1=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}, \\
\cdots \ldots \ldots . .
\end{array}
$$

Also, 1 can be expressed as the sum of unit fractions with odd denominators in at least nine terms, for example:
$$
1=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{35}+\frac{1}{45}+\frac{1}{231} \text {, }
$$

There are also four other solutions, the first six terms of which are the same as above, and the denominators of the last three terms are $(21,135,10395)$, $(33,45,385)$, $(21,165,693)$, $(21,231,315)$.
If the sum is to have the smallest maximum denominator of the simplest fraction, it can be expressed as
$$
1=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{33}+\frac{1}{35}+\frac{1}{45}+\frac{1}{55}+\frac{1}{77}+\frac{1}{105} .
$$

Note 2: If the following conclusion is known, solving this example will be easier.
Decompose 1 into the sum of some unit fractions, i.e.,
$$
1=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}},
$$

If there exists a prime $p$ and a positive integer $k$, such that $p^{k+1} \nmid a_{i}(1 \leqslant i \leqslant n)$, but there is some $i_{0}$, such that $p^{k} \mid a_{i_{0}}$, then there must be another $j \neq i_{0}, \quad 1 \leqslant j \leqslant n$, such that $p^{k} \mid a_{j}$.

Proof: If the proposition is not true, then $p^{k} \nmid a_{j}$ for any $j \neq i_{0}$. Then, multiplying both sides of $\circledast$ by an appropriate multiple of $p^{k-1}$, the right side of $\circledast$ except the $\frac{1}{a_{i_{0}}}$ term would all be integers, while the left side is an integer, which is impossible.
Therefore, the proposition is true.
By the above proposition, considering the prime $p=5, k=2$, we can eliminate $(B),(D)$; considering the prime $p=3$, $k=2$, we can eliminate $(A)$.
Therefore, the answer is $(C)$.