2005 AMC 8 Problems/Problem 24

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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

It is easier to work backwards from $200$, and have keys that display [-1] and [x0.5]. Use the second key when the number is even, and the first key when the number is odd until you get one. We get:

\[200, 100, 50, 25, 24, 12, 6, 3, 2, 1\]

This takes $\boxed{\textbf{(B)}\ 9}$ keystrokes.

See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
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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