2018 AMC 10B Problems/Problem 13: Difference between revisions

Revision as of 08:32, 17 February 2018

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
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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 ==Solution 2==
-If we divide each number by <math>101</math>, we see a pattern occuring in every 4 numbers. <math>101, 1000001, 10000000001, \dots. We divide </math>2018<math> by </math>4<math> to get </math>504<math> with </math>2<math> left over. One divisible number will be in the </math>2<math> left over, so out answer is </math>\boxed{\textbf{(C) } 505}$.
+If we divide each number by <math>101</math>, we see a pattern occuring in every 4 numbers. <math>101, 1000001, 10000000001, \dots</math>. We divide <math>2018</math> by <math>4</math> to get <math>504</math> with <math>2</math> left over. One divisible number will be in the <math>2</math> left over, so out answer is <math>\boxed{\textbf{(C) } 505}</math>.
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2018 AMC 10B Problems/Problem 13: Difference between revisions

Revision as of 08:32, 17 February 2018

Contents

Problem

Solution

Solution 2

See Also