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

(Solution 2)
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==Solution 2==
 
==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>.
  
 
==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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