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 press the button that multiplies by 2 the most, but we are going down instead of up, so we divide by 2 instead. If we come across an odd number, we will subtract that number by 1.

=,

$100/2$ = $50$ , $50/2$ = $25$ , $25-1$ = $24$ , $24/2$ = $12$ , $12/2$ = $6$ , $6/2$ = $3$ , $3-1$ = $2$ , and $2/2$ = $1$ . We made our way down to 1 but since it is the amount of times the button is pressed then 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
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2005 AMC 8 Problems/Problem 24

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