9. 

Calculate the number $8^{2021}$, find the sum of the digits in this number, and write down the result. Then, in the newly written number, find the sum of the digits and write down the result again. These actions were repeated until a single-digit number was obtained. Find this number.
# Solution.

Consider the natural powers of 8. Notice that even powers of the number 8 give a remainder of 1 when divided by 9, while odd powers (including the number \(8^{2021}\)) give a remainder of 8. Indeed, let's analyze the powers of 8:

\[
\begin{gathered}
8^{2}=(9-1)^{2}=9 n+1, n \in N \\
8^{3}=(9 n+1) \cdot 8=9 k+8, k \in N \\
8^{4}=(9 k+8) \cdot 8=9 k \cdot 8+8^{2}=9 k \cdot 8+9 n+1=9 \cdot s+1, s \in N
\end{gathered}
\]

Assuming that \(8^{2 m-1}=9 t+8, t \in N, m \in N, m \geq 2\), we get:

\[
\begin{aligned}
& 8^{2 m}=8^{2 m-1} \cdot 8=(9 t+8) \cdot 8=9 t \cdot 8+8^{2}=9 t \cdot 8+9 n+1=9 q+1, q \\
& \quad \in N
\end{aligned}
\]

\[
8^{2 m+1}=8^{2 m} \cdot 8=(9 q+1) \cdot 8=9 q \cdot 8+8
\]

Since an integer when divided by 9 gives the same remainder as the sum of its digits, the sum of the digits of the number \(8^{2021}\) and the sum of the digits of subsequent summation results, when divided by 9, have the same remainder of 8.

Answer. 8.