# 2005 AMC 8 Problems/Problem 24

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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 $10-1$= $\boxed{\textbf{(B)}\ 9}$! - $\boxed{\textbf{Javapost}}$

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