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| − | ==Problem==
| + | #REDIRECT [[2013 AMC 12B Problems/Problem 7]] |
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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>.
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| − | == See also ==
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| − | {{AMC10 box|year=2013|ab=B|num-b=12|num-a=14}}
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