2005 AMC 8 Problems/Problem 24

Problem

A certain calculator has only two keys [+1] and [x2]. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed "9" and you pressed [+1], it would display "10." If you then pressed [x2], it would display "20." Starting with the display "1," what is the fewest number of keystrokes you would need to reach "200"?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution

First we can start at 200 and work our way down to 1. Since we want to do divide by two the most, so if we come across an odd number we just subtract 1. So $200/2$ = $100$ , $100/2$ = $50$ , $50/2$ = $25$ , $25-1$ = $24$ , $24/2$ = $12$ , $12/2$ = $6$ , $6/2$ = $3$ , $3-1$ = $2$ , and $2/2$ = $1$ . We made our way back to 1 but since it is the amount of times the button is pressed than the answer should be $10-1$ = $\boxed{\textbf{(B)}\ 9}$ . -$\boxed{\textbf{Javapost}

2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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 AJHSME/AMC 8 Problems and Solutions

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2005 AMC 8 Problems/Problem 24

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