All About 2018 -- C

Let the $n$ th term of the sequence $(101,1001,10001,100001\cdots)$ be $a_n$ , then $a_n$ is given by $10^{n+1}+1$ for $n\in[1,2018]$ .

So, we want $n$ such that: $10^{n+1}+1\equiv0 \mod (101)$

$\Rightarrow 10^{n+1}\equiv-1 \mod (101)$

$\Rightarrow$ $100^{\frac{n+1}{2}}\equiv-1 \mod (101)$

$\Rightarrow$ $(-1)^{\frac{n+1}{2}}\equiv-1 \mod (101).$

This happens at $\frac{(n+1)}{2}\equiv1 \mod (2).$ which is equivalent to $n\equiv1 \mod (4)$ So the numbers are $a_1,a_5,a_9\cdots a_{2017}$ .

Now, let these numbers form another sequence $b_n$ i.e. $b_1=a_1,b_2=a_5\cdots,b_{505}=a_{2017}$

The number of terms in this sequence is $505$ . This means that from the first $2018$ number of the given sequence, $\boxed{505}$ numbers are divisible by $101$

We note that the $n$ th member of the sequence is given by $a_n = 10^{n+1}+1$ for $n \in \mathbb N$ . We need to find the number of $m$ such that $10^m + 1 \equiv 0 \text{ (mod 101)}$ for $2 \le m \le 2019$ .

$\begin{aligned} 10^m + 1 & \equiv 0 \text{ (mod 101)} & \small \color{#3D99F6} \text{for }2 \le m \le 2019 \\ 10^m & \equiv - 1 \equiv 100 \text{ (mod 101)} \\ 10^{m}-10^2 & \equiv 0 \text{ (mod 101)} \\ 100(10^{m-2}-1) & \equiv 0 \text{ (mod 101)} \end{aligned}$

Now we note that $10^{m-2}-1 = 0, 9, 99, 999, ... \underbrace{999...999}_{\text{\# of 9 = 2017}}$ , where $0, \underbrace{9999}_{\text{\#of 9=4}}, \underbrace{99999999}_{\text{\#of 9=8}}, ... \underbrace{999...999}_{\text{\#of 9=2016}}$ are divisible by 101. Therefore there are $\dfrac {2016}4 + 1=\boxed{505}$ number of $m$ satisfy the condition.