# 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}\$.

## See Also

 2018 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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