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

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(Redirected page to 2013 AMC 12B Problems/Problem 3)
 
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==Problem==
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#REDIRECT [[2013 AMC 12B Problems/Problem 3]]
When counting from <math>3</math> to <math>201</math>, <math>53</math> is the <math>51^\mathrm{st}</math> number counted. When counting backwards from <math>201</math> to <math>3</math>, <math>53</math> is the <math>n^\mathrm{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>
 
 
 
==Solution==
 
 
 
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.
 
== See also ==
 
{{AMC10 box|year=2013|ab=B|num-b=3|num-a=5}}
 

Latest revision as of 12:09, 7 April 2013