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

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 (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 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 (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 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.

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