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

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{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #3]] and [[2013 AMC 10B Problems|2013 AMC 10B #4]]}}
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==Problem==
 
==Problem==
 
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>?
 
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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==Solution==
 
==Solution==
Note that <math>n</math> is equal to the number of integers between <math>53</math> and <math>201</math>, inclusive. Thus, <math>n=201-53+1=\boxed{\textbf{(D)}149}</math>
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Note that <math>n</math> is equal to the number of integers between <math>53</math> and <math>201</math>, inclusive. Thus, <math>n=201-53+1=\boxed{\textbf{(D)}\ 149}</math>
  
 
== 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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{{MAA Notice}}

Latest revision as of 17:19, 12 June 2018

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 $51^{st}$ number counted. When counting backwards from $201$ to $3$, $53$ is the $n^{th}$ 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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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