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

Revision as of 11:18, 7 April 2013

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, $n=201-53+1=\boxed{\textbf{(D)}\ 149}$

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

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 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

@@ Line 11: / Line 11: @@
 == See also ==
 {{AMC12 box|year=2013|ab=B|num-b=2|num-a=4}}
+{{AMC10 box|year=2013|ab=B|num-b=3|num-a=5}}

Page

Toolbox

Search

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

Revision as of 11:18, 7 April 2013

Problem

Solution

See also