## Task 17/84

Let $a$ be a 1984-digit natural number that is divisible by 3. Let $b$ be the sum of the digits of $a$, $c$ the sum of the digits of $b$, and $d$ the sum of the digits of $c$. What values can $d$ take?
If a number is divisible by 3, then its cross sum is also divisible by 3. Therefore, with a and b, $\mathrm{c}$ and d being divisible by 3. Clearly, the following estimates apply to b, c, and d:

$$
b \leq 9 \cdot 1984=17856 ; \quad c<9+9+9+9=36 ; \quad d<2+9=11
$$

Thus, for $\mathrm{d}$, only the values $d_{1}=3 ; d_{2}=6 ; d_{3}=9$ are possible.