# 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 1 (Unrigorous)

We can start at $200$ and work our way down to $1$. We want to press the button that multiplies by $2$ the most, but since we are going down instead of up, we divide by $2$ instead. If we come across an odd number, then we will subtract that number by $1$. Notice

$200 \div 2 = 100$,
$100 \div 2 = 50$,
$50 \div 2 = 25$,
$25-1 = 24$,
$24 \div 2 = 12$,
$12 \div 2 = 6$,
$6 \div 2 = 3$,
$3-1 = 2$,
$2 \div 2 = 1$.


Since we've reached $1$, it's clear that the answer should be $\boxed{\textbf{(B)}\ 9}$- $\boxed{\textbf{Javapost}}$.

## Solution 2 (Bounding) - ike.chen

Clearly, there exists a construction for $9$ keystrokes, as shown above. Now, we show this is the smallest possible number of keystrokes.

If there are at most $7$ keystrokes, then the highest number we can reach is $128 < 200$.

If there are $8$ keystrokes, then we consider the following cases:

- $8$ [x2]: This will clearly result in $256$, which isn't desired.

- $7$ [x2], $1$ [+1]: The two largest numbers we can reach from this case are $256$ and $192$, so we know this combination will not work.

- If we use at most $6$ repetitions of [x2], then our number will be at most $(1 + 2) \cdot 2^{6} = 192$, so all of these combinations are bad.

Hence, $\boxed{\textbf{(B)}\ 9}$ is the answer.

## See Also

 2005 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 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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