2. Consider the number $\mathrm{a}_{\mathrm{n}}=18 \underbrace{77 \ldots 7}_{\text {n times }} 889$, where $n$ is a natural number, and $\mathrm{C}_{\mathrm{n}}$ is the quotient of the division of the number $\mathrm{a}_{\mathrm{n}}$ by 13.

a) Show that $\mathrm{a}_{\mathrm{n}}$ is divisible by 13 for any $n$.

b) Determine $n$ for which $s\left(a_{n}\right)=2 s\left(c_{n}\right)$, where $s(m)$ represents the sum of the digits of the number $m$.
Solution:

a) $a_{n}=18 \cdot 10^{n+3}+7 \cdot 10^{3} \cdot\left(10^{n-1}+10^{n-2}+\cdots+10+1\right)+889=13 \cdot 10^{n+3}+13 \cdot 68+$

$7 \cdot 10^{3} \cdot \frac{10^{n}-1}{9}+5 \cdot 10^{n+3}+5$.

However, $7 \cdot 10^{3} \cdot \frac{10^{n}-1}{9}+5 \cdot 10^{n+3}+5=\frac{52 \cdot 10^{n+3}-6955}{9}=\frac{13 \cdot\left(4 \cdot 10^{n+3}-535\right)}{9}=\frac{13}{9} \cdot \overline{\underbrace{399 \ldots 9465}_{n+4 \text { digits }}}=$

$13 \cdot \underbrace{44 \ldots 4385}_{n+3 \text { digits }}$. Therefore, $a_{n}=13 \cdot(10^{n+3}+68+\underbrace{\overline{44 \ldots 485}}_{n+3 \text { digits }})$, so $a_{n}$ is divisible by 13.

4 points

b) $s\left(a_{n}\right)=1+8+7 n+8+8+9=7 n+34$

$\mathrm{C}_{n}=\underbrace{\overline{144 \ldots 4453}}_{n+4 \text { digits }} \Rightarrow \mathrm{s}\left(\mathrm{C}_{\mathrm{n}}\right)=9+4 \cdot(n+1)=4 n+13 \Rightarrow 7 n+34=2 \cdot(4 n+13) \Rightarrow n=8$

3 points