$1 \cdot 15$ Choose a 1962-digit number that is 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 $c$?
[Solution] Since any number and the sum of its digits have the same remainder when divided by 9,
thus $c \geqslant 9$, and $9 \mid c$.
On the other hand, $a \leqslant 1962 \times 9 < 19999$,
therefore $b < 1 + 4 \times 9 = 37$,
so $c \leqslant 11$. Hence, $c=9$.