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 (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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