## SUBJECT 3

Let the number $\mathrm{n}=9+99+999+\ldots+\underbrace{99 \ldots 9}_{\text {2016 digits }}+2016$.

a) Show that the number n is divisible by 90.

b) Find the quotient and remainder of the division of $\mathrm{n}$ by 11111.
## SUBJECT 3

$$
\begin{aligned}
& \text { a) } n=10-1+100-1+1000-1+\cdots+1 \underbrace{00 \ldots 0}-1+2016 \ldots \ldots \ldots 1 p \\
& 2016 \text { zeros } \\
& =10+100+1000+\cdots+\underbrace{100 \ldots 0}_{2016 \text { zeros }}-\underbrace{(1+1+1+\cdots 1)}_{2016 \text { digits } 1}+2016=
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_3480621dd5c5c8cebb70g-2.jpg?height=68&width=1259&top_left_y=1522&top_left_x=197)

$$
\begin{aligned}
& \text { 111...10 } \vdots 10
\end{aligned}
$$

Since the sum of the digits is 2016, which is divisible by 9, and $(9 ; 10)=1 \Rightarrow n \vdots 90 \ldots \ldots . .1 p$

b) $n=\underbrace{111 \ldots 10}_{2016 \text { digits } 1} 0=11111 \cdot 10^{2012}+11111 \cdot 10^{2007}+\cdots+11111 \cdot 10^{5 \mathrm{k}+2}+$ $+\cdots+11111 \cdot 10^{2}+10=$ $1 p$

$=11111 \cdot c+10 \Rightarrow r=10$

$1 p$

$c=10^{2012}+10^{2007}+\cdots+10^{2}$ $1 p$