Ttykuu

This is a problem from the AMC 8 (math contest):

A certain calculator has only two keys $[+1]$ and $[times 2]$. 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 $[times 2]$, it would display “$20$”. Starting with the display “$1$”, what is the fewest number of keystrokes you would need to reach “$200$”?

Intuitively I worked back from $200$, dividing by $2$ until I reached an odd number, subtracting $1$ when I did, etc..to reach the correct answer of $9$ steps.

However, I can't figure out how to convince myself beyond any doubt that it is the optimal solution. In other words, I can't prove it mathematically.
The best I can come up with is that beyond the first step from $1$ to $2$, multiplication by $2$ is always going to yield a larger step than addition by $1$ and therefore I should take as many $[times 2]$ steps as I can. This doesn't feel rigorous enough, though.

EDIT: Just to be clear, I am asking for a proof or at least more rigorous explanation of why this is the optimal solution.

Setting the display to binary base, $[times2]$ inserts a $0$ to the right and $[+1]$ increments; if the rightmost digit is a zero, it just turns it to a $1$.

Using these rules you build a number of $o$ ones and $z$ zeroes in $o-1+z$ keystrokes, starting from $1$. This seems close to optimal.

Look at what the operations $+$ and $times$ do to the binary expansion of a number:

• 
  $times$ appends a $0$, and increases the length by one, leaving the total number of $1$'s unchanged;


• if the final digit is $0$, then $+$ increases the number of $1$'s by one, but doesn't change the length;


• if the final digit is $1$, then $+$ doesn't increase the total number of $1$'s (it may in fact decrease it), and doesn't increase the total length by more than $1$.

The binary expansion of $200$ is $200_10=11001000_2$.

This has three $1$'s, so to get there from a starting point of $1$, we need at least two operations of the second kind.

And it has a length of eight bits, so we need at least seven operations of the first or the third kind.

So the minimum number of operations is $9$, as required.

You can proceed by induction on $n$, showing that the quickest way to reach any even number $2n$ involves doubling on the last step, which is clearly true for the base case $n=1$ (where doubling and adding $1$ have a tomato-tomahto relationship).

Now if the last step to reach $2n+2$ isn't doubling, it can only be adding $1$ from $2n+1$. But $2n+1$ can only be reached by adding $1$ from $2n$, at which point the inductive hypothesis says the next previous number was $n$. But you can get from $n$ to $2n+2$ more quickly in two steps: add $1$, then double. So the last step in the quickest route to $2n+2$ is doubling from $n+1$.

I'll make a try

Since $200=2^7+2^6+2^3$ you will need at least $8$ steps to reach $200$ (since we start from $1$ and we get a number of the form $2^a+...+2^l$) so it remains to show that $8$ steps are not enough.

Maybe you could try to show that if there was a solution with $8$ steps then it would contain only one $+1$ which contradicts the fact that in $200=2^7+2^6+2^3$ we have two $+$