4. let $q(n)$ be the sum of the digits of the natural number $n$. Determine the value of

$$
q\left(q\left(q\left(2000^{2000}\right)\right)\right)
$$

## Solution
It applies $q\left(2000^{2000}\right)=q\left(2^{2000}\right)$, because the two numbers only differ by appended zeros. We now estimate. The following applies $2^{2000}=4 \cdot 8^{666}<10^{667}$ and therefore

$$
q\left(2^{2000}\right) \leq 9 \cdot 667=6003
$$

Furthermore

$$
q\left(q\left(2^{2000}\right)\right) \leq 5+9+9+9=32 \quad \text { and } \quad q\left(q\left(q\left(2^{2000}\right)\right)\right) \leq 2+9=11
$$

On the other hand, $q(n) \equiv n(\bmod 9)$ applies to all natural numbers $n$. With $\varphi(9)=6$ this results in

$$
2000^{2000} \equiv 2^{333 \cdot 6+2}=\left(2^{6}\right)^{333} \cdot 4 \equiv 4 \quad(\bmod 9)
$$

The number you are looking for is therefore $\leq 11$ on the one hand and $\equiv 4(\bmod 9)$ on the other, i.e. equal to 4 .