2013 AMC 12B Problems/Problem 7

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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 $\boxed{\textbf{(E) }8}$.

See also

2013 AMC 12B (ProblemsAnswer KeyResources)
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