2018 AMC 10B Problem 13

The $m$-th term is $10^{m+1} + 1$, so the first $2018$ terms are $10^k + 1$ for $k = 2, 3, \dots, 2019$.

Work modulo $101$. Since $10^2 = 100 \equiv -1$, the powers of $10$ cycle with period $4$:
$$10^1 \equiv 10,\quad 10^2 \equiv -1,\quad 10^3 \equiv -10,\quad 10^4 \equiv 1,\quad \dots$$
We need $10^k \equiv -1 \pmod{101}$, which happens exactly when $k \equiv 2 \pmod 4$.

Count the exponents $k \in \{2, \dots, 2019\}$ with $k \equiv 2 \pmod 4$: they are $2, 6, 10, \dots, 2018$, an arithmetic progression with common difference $4$, so there are
$$\frac{2018 - 2}{4} + 1 = 504 + 1 = 505$$
of them.

The answer is $\boxed{\textbf{(C)}\ 505}$.

Divisibility of $10^k + 1$ by a fixed number depends only on $k$ modulo the order of $10$; here $10^2 \equiv -1$ makes the order $4$ and singles out $k \equiv 2$. Translate the problem into a congruence condition on the index, then count indices carefully, including the offset in where the sequence starts.

Check the first few terms directly: $101$ is divisible, $1001 = 7 \cdot 11 \cdot 13$ is not, $10001 = 73 \cdot 137$ is not, $100001 = 11 \cdot 9091$ is not, $1000001 = 101 \cdot 9901$ is. Every fourth term starting with the first works, and among $2018$ terms that is $\lceil 2018/4 \rceil = 505$.

• Indexing the exponent from 1 instead of 2 (the first term is 10^2 + 1), or counting every other term (1009) as divisible instead of every fourth.
• Getting $504$ by computing $2018/4$ and truncating, forgetting that the pattern starts at the very first term.