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 ...")
 
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==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 \>
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<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8</math>
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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?
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<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8 </math>
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==Solution==
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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 16:49, 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 $\boxed{\textbf{(E) }8}$.