# 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

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.

$200/2$=$100$,
$100/2$=$50$,
$50/2$=$25$,  $25-1$=$24$,
$24/2$=$12$,
$12/2$=$6$,
$6/2$=$3$,  $3-1$=$2$,
and
$2/2$=$1$.


We made our way down to 1 but since it is the amount of times the button is pressed then the answer should be $10-1$=$\boxed{\textbf{(B)}\ 9}$! - $\boxed{\textbf{Javapost}}$

## 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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