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

(Created page with "==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 <m...")
 
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<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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==Solution==
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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>

Revision as of 15:38, 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, $n=201-53+1=\boxed{\textbf{(D)}149}$

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