Difference between revisions of "2013 AMC 10B Problems/Problem 4"

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==Solution==
 
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Notice that for a number <math>n \le 201</math>, the place in which it is counted is the same as <math>201 - n + 1 = 202 - n</math>.  (This is evident due to the fact that 201 is the 1st number counted, and every number after that increments the place by one).  Thus, 53 is the <math>202 - 53 = \boxed{\textbf{(D) }149}</math>th number counted.

Revision as of 16:24, 21 February 2013

Problem

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

Notice that for a number $n \le 201$, the place in which it is counted is the same as $201 - n + 1 = 202 - n$. (This is evident due to the fact that 201 is the 1st number counted, and every number after that increments the place by one). Thus, 53 is the $202 - 53 = \boxed{\textbf{(D) }149}$th number counted.