9.6 Take any 1962-digit number divisible by 9, and let the sum of its digits be $a$, the sum of the digits of $a$ be $b$, and the sum of the digits of $b$ be $c$. What is $c$?
(2nd All-Russian Mathematical Olympiad, 1962)
［Solution］From the given conditions,
$$
\begin{array}{l}
a \leqslant 1962 \cdot 9=17658<19999, \\
b \leqslant 1+4 \cdot 9=37, \\
c \leqslant 3+9=12 .
\end{array}
$$

Since the sum of the digits of any positive integer has the same remainder when divided by 9 as the integer itself.
Given that the 1962-digit number is divisible by 9, then $a, b, c$ are all divisible by 9.
Given that
$$
c \leqslant 12 \text {, }
$$

thus
$$
c=9 \text {. }
$$