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}$.

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2018 AMC 10B Problems/Problem 13

Contents

Problem

Solution

Solution 2

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