Answer :
The number of such n is [tex]$\boxed{2}$.[/tex]
The first term of the sequence is [tex]$101$.[/tex]
Therefore, the $n$th term is given by [tex]$10^n+1$.[/tex]
We must determine how many of the first $2018$ numbers in the sequence are divisible by [tex]$101$.[/tex]
By the Remainder Theorem, the remainder when $10^n+1$ is divided by $101$ is $10^n+1 \mod 101$.
We must find all values of $n$ between $1$ and $2018$ such that
[tex]$10^n+1 \equiv 0 \mod 101$.[/tex]
By rearranging this equation, we have [tex]$$10^n \equiv -1 \mod 101.$$[/tex]
Notice that
[tex]$10^0 \equiv 1 \mod 101$, \\$10^1 \equiv 10 \mod 101$, \\$10^2 \equiv -1 \mod 101$, \\$10^3 \equiv -10 \mod 101$, \\$10^4 \equiv 1 \mod 101$[/tex]
, and so on.
Thus, the remainder of the powers of $10$ alternate between 1 and -1.
Since $2018$ is even, we must have [tex]$10^{2018} \equiv 1 \mod 101$.[/tex]
Therefore, we have [tex]$$10^n \equiv -1 \mod 101$[/tex] if and only if n is an odd multiple of $1009$ and $n$ is less than or equal to 2018.
The number of such n is [tex]$\boxed{2}$.[/tex]
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