2005 AMC 8 Problems/Problem 24

Revision as of 15:04, 28 September 2020 by Javapost (talk | contribs) (Solution)

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. $$ (Error compiling LaTeX. Unknown error_msg)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 (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AJHSME/AMC 8 Problems and Solutions

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