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

Revision as of 18:36, 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 $\boxed{\textbf{(C) } 505}$ .

Solution 3

Note that $909$ is divisible by $101$ , and thus $9999$ is too. We know that $101$ is divisible and $1001$ isn't so let us start from $10001$ . We subtract $9999$ to get 2. Likewise from $100001$ we subtract, but we instead subtract $9999$ times $10$ or $99990$ to get $11$ . We do it again and multiply the 9's by $10$ to get $101$ . Following the same knowledge, we can use mod $101$ to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is ${0,-9,-99 ( 2),11, 0, ...}$ . Thus it repeats every four. Consider the sequence after the 1st term and we have 2017 numbers. Divide $2017$ by four to get $504$ remainder $1$ . Thus the answer is $504$ plus the 1st term or $\boxed{\textbf{(C) } 505}$ .

-googleghosh

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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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</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>.
+==Solution 3==
+Note that <math>909</math> is divisible by <math>101</math>, and thus <math>9999</math> is too. We know that <math>101</math> is divisible and <math>1001</math> isn't so let us start from <math>10001</math>. We subtract <math>9999</math> to get 2. Likewise from <math>100001</math> we subtract, but we instead subtract <math>9999</math> times <math>10</math> or <math>99990</math> to get <math>11</math>. We do it again and multiply the 9's by <math>10</math> to get <math>101</math>. Following the same knowledge, we can use mod <math>101</math> to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is <math>{0,-9,-99 ( 2),11, 0, ...}</math>. Thus it repeats every four. Consider the sequence after the 1st term and we have 2017 numbers. Divide <math>2017</math> by four to get <math>504</math> remainder <math>1</math>. Thus the answer is <math>504</math> plus the 1st term or <math>\boxed{\textbf{(C) } 505}</math>.
+-googleghosh
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