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"?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$
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 $\boxed{\textbf{(B)}\ 9}$- $\boxed{\textbf{Javapost}}$. Because we only subtracted $1$ when we had to, this is optimal. ~Roy2020