Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AMC 12B Problems/Problem 7"
(9 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #7]] and [[2013 AMC 10B Problems|2013 AMC 10B #13]]}}
 +
 
==Problem==
 
==Problem==
 +
Jo and Blair take turns counting from <math>1</math> to one more than the last number said by the other person. Jo starts by saying <math>``1"</math>, so Blair follows by saying <math>``1, 2"</math>. Jo then says <math>``1, 2, 3"</math>, and so on. What is the <math>53^{\text{rd}}</math> number said?<br \>
 +
 +
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8</math>
  
−
Jo and Blair take turns counting from <math>1</math> to one more than the last number said by the other person. Jo starts by saying "<math>1</math>", so Blair follows by saying "<math>1, 2</math>" . Jo then says "<math>1, 2, 3</math>" , and so on. What is the <math>53^{rd}</math> number said?
+
==Solution==
  
−
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8 </math>
+
We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc.  Thus, after nine "turns", <math>1+2+3+4+5+6+7+8+9=45</math> numbers have been said.  In the tenth turn, the eighth number will be the 53rd number said, because <math>53-45=8</math>.  Since we are starting from 1 every turn, the 53rd number said will be <math>\boxed{\textbf{(E) }8}</math>.
  
−
==Solution==
+
==Video Solution==
 +
https://youtu.be/a-3CAo4CoWc
  
−
We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc.  Thus, after nine "turns," <math>1+2+3+4+5+6+7+8+9=45</math> numbers have been said.  In the tenth turn, the eighth number will be the 53rd number said, as <math>53-45=8</math>.  Since we're starting from 1 each time, the 53rd number said will be <math>\boxed{\textbf{(E) }8}</math>.
+
~[[User:icematrix2|icematrix2]]
  
 
== See also ==
 
== See also ==
 +
{{AMC10 box|year=2013|ab=B|num-b=12|num-a=14}}
 
{{AMC12 box|year=2013|ab=B|num-b=6|num-a=8}}
 
{{AMC12 box|year=2013|ab=B|num-b=6|num-a=8}}
 +
{{MAA Notice}}

Revision as of 22:08, 30 September 2020

The following problem is from both the 2013 AMC 12B #7 and 2013 AMC 10B #13, so both problems redirect to this page.

Problem

Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$, so Blair follows by saying $``1, 2"$. Jo then says $``1, 2, 3"$, and so on. What is the $53^{\text{rd}}$ number said?

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$

Solution

We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns", $1+2+3+4+5+6+7+8+9=45$ numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, because $53-45=8$. Since we are starting from 1 every turn, the 53rd number said will be $\boxed{\textbf{(E) }8}$.

Video Solution

https://youtu.be/a-3CAo4CoWc

~icematrix2

See also

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 by
Problem 6 		Followed by
Problem 8
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_7&oldid=134419"