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"

(Created page with "==Problem== Jo and Blair take turns counting from 1 to one more than the last number said by the other person. Jo starts by saing "1", so Blair follows by saying "1, 2". Jo then ...")
 
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
Jo and Blair take turns counting from 1 to one more than the last number said by the other person. Jo starts by saing "1", so Blair follows by saying "1, 2". Jo then says "1, 2, 3", and so on. What is the <math>53^{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?
 +
 
 +
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8 </math>
 +
 
 +
==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," <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>.

Revision as of 15:49, 22 February 2013

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 $53^{rd}$ 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, as $53-45=8$. Since we're starting from 1 each time, the 53rd number said will be $\boxed{\textbf{(E) }8}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_7&oldid=51400"