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Difference between revisions of "2013 AMC 12B Problems/Problem 3"

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==Problem==
 
==Problem==
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When counting from <math>3</math> to <math>201</math>, <math>53</math> is the <math>51^{st}</math> number counted. When counting backwards from <math>201</math> to <math>3</math>, <math>53</math> is the <math>n^{th}</math> number counted. What is <math>n</math>?
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When counting from <math>3</math> to <math>201</math>, <math>53</math> is the <math>51^{\text{st}}</math> number counted. When counting backwards from <math>201</math> to <math>3</math>, <math>53</math> is the <math>n^{\text{th}}</math> number counted. What is <math>n</math>?
  
 
<math>\textbf{(A)}\ 146 \qquad \textbf{(B)}\ 147 \qquad \textbf{(C)}\ 148 \qquad \textbf{(D)}\ 149 \qquad \textbf{(E)}\ 150</math>
 
<math>\textbf{(A)}\ 146 \qquad \textbf{(B)}\ 147 \qquad \textbf{(C)}\ 148 \qquad \textbf{(D)}\ 149 \qquad \textbf{(E)}\ 150</math>
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== See also ==
 
== See also ==
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{{AMC10 box|year=2013|ab=B|num-b=3|num-a=5}}
 
{{AMC12 box|year=2013|ab=B|num-b=2|num-a=4}}
 
{{AMC12 box|year=2013|ab=B|num-b=2|num-a=4}}
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{{AMC10 box|year=2013|ab=B|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 17:47, 15 October 2024

The following problem is from both the 2013 AMC 12B #3 and 2013 AMC 10B #4, so both problems redirect to this page.

Problem

When counting from $3$ to $201$, $53$ is the $51^{\text{st}}$ number counted. When counting backwards from $201$ to $3$, $53$ is the $n^{\text{th}}$ number counted. What is $n$?

$\textbf{(A)}\ 146 \qquad \textbf{(B)}\ 147 \qquad \textbf{(C)}\ 148 \qquad \textbf{(D)}\ 149 \qquad \textbf{(E)}\ 150$

Solution

Note that $n$ is equal to the number of integers between $53$ and $201$, inclusive. Thus, $n=201-53+1=\boxed{\textbf{(D)}\ 149}$

See also

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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
2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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 12 Problems and Solutions

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

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