10.57 Given a 29-digit number $x=\overline{a_{1} a_{2} \cdots a_{28} a_{29}}\left(0 \leqslant a_{k} \leqslant 9, \quad a_{1} \neq 0, k=1,2, \cdots, 29\right)$. It is known that for each $k$, the digit $a_{k}$ appears $a_{30-k}$ times in the representation of the number (for example, if $a_{10}=7$, then the digit $a_{20}$ appears 7 times). Find the sum of the digits of $x$.
(34th Moscow Mathematical Olympiad, 1971)
[Solution] Let $k(A)$ denote the number of occurrences of digit $A$.
Then, from the given conditions, if digits $B, C$ are in symmetric positions, we have $k(B) = C, k(C) = B$. For example, if $a_{14}=5$, then $a_{16}$ appears 5 times, i.e., $k\left(a_{16}\right)=a_{14}=5$. If $a_{16}=7$, then $a_{14}$ appears 7 times, i.e., $k\left(a_{14}\right)=a_{16}=7$. This means that these digits should always be in symmetric positions. Therefore, if $k(B)=k(C)$ and $B=C$, it can be immediately deduced that $k(A)=A$ always holds, and the digit in the middle position (i.e., the 15th position) should appear an odd number of times, while the other digits should appear an even number of times.
Thus, the number of digits in $x$ does not exceed
\[
\begin{aligned}
& k(2)+k(4)+k(6)+k(8)+k(9) \\
= & 2+4+6+8+9 \\
= & 29 .
\end{aligned}
\]

Based on the conditions, we know that there are exactly 29 digits.
This means that $x$ has exactly 2 occurrences of 2, 4 occurrences of 4, 6 occurrences of 6, 8 occurrences of 8, and 9 occurrences of 9.
Thus, the sum of the digits of $x$ is
\[
2^{2}+4^{2}+6^{2}+8^{2}+9^{2}=201.
\]