# Difference between revisions of "2018 AMC 10B Problems/Problem 13"

## Problem

How many of the first $2018$ numbers in the sequence $101, 1001, 10001, 100001, \dots$ are divisible by $101$?

$\textbf{(A) }253 \qquad \textbf{(B) }504 \qquad \textbf{(C) }505 \qquad \textbf{(D) }506 \qquad \textbf{(E) }1009 \qquad$

## Solution

Note that $10^{2k}+1$ for some odd $k$ will suffice $\mod {101}$. Each $2k \in \{2,4,6,\dots,2018\}$, so the answer is $\boxed{\textbf{(C) } 505}$ (AOPS12142015)

## Solution 2

If we divide each number by $101$, we see a pattern occuring in every 4 numbers. $101, 1000001, 10000000001, \dots$. We divide $2018$ by $4$ to get $504$ with $2$ left over. One divisible number will be in the $2$ left over, so out answer is $\boxed{\textbf{(C) } 505}$.