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

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

Revision as of 09:32, 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$\boxed{\textbf{(C) } 505}$.

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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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