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

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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>
 
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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== See also ==
 +
{{AMC12 box|year=2013|ab=B|num-b=2|num-a=4}}

Revision as of 17:02, 22 February 2013

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 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
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