How can I prove that starting from $0$, and repeatedly taking $+10,-10,\times10$ or $\div10$, requires at least $9$ operations to get to $2019$? Sometimes, I teach my sister math homework. Today, when she was doing her math problem solving summer homework, I teached her how to do the problems except the final problem that I don't know how to do.

A machine is showing the number $0$. It has $4$ buttons: $\boxed{+10}, \boxed{-10}, \boxed{×10},$ and $\boxed{÷10}$. If you press a button, the number that the machine is showing will be changed by the operation on the button. At least how many times do you need to press the buttons in order to make $2019$? 

I think that the answer is 9, and the sequence is $\boxed{+10}, \boxed{+10}, \boxed{×10},\boxed{×10},\boxed{+10}, \boxed{+10}, \boxed{×10},\boxed{-10},\boxed{÷10}$.
But is it correct and how can I explain to my sister that it is the least possible value?
 A: You can say that every button $\boxed{+10}$ or $\boxed{-10}$ has a different worth depending on how many buttons $\boxed{\times 10}$ or $\boxed{÷10}$ are pressed at any point after it. Each $\boxed{\times 10}$ increases the values of any previous $\boxed{+10}$ or $\boxed{-10}$ by a factor of $10$, and each $\boxed{÷10}$ decreases the values of any previous $\boxed{+10}$ or $\boxed{-10}$ by a factor of $10$.
For example, if after you press $\boxed{+10}$ and then sometimes after you press $\boxed{\times 10}$ three times, and $\boxed{÷10}$ one time, the original $\boxed{+10}$ will at the end be worth $10\times 10\times 10 \times 10 ÷10 = 1000$.
Therefore, in the end, each pressing of $\boxed{+10}$ may be worth $10$, $100$, $1000$... or $1$, $0.1$, $0.01$... depending on how many $\boxed{\times 10}$ and $\boxed{÷10}$ are pressed after them. Similarily, $\boxed{-10}$ may be worth $-10$, $-100$, $-1000$... or $-1$, $-0.1$, $-0.01$...
The required number needs to be constructed from these final worths of $\boxed{+10}$ and $\boxed{-10}$. The easiest way to write $2019$ in this way is $$ 2019 = 1000 + 1000 +10 + 10 -1$$
That means you need:


*

*Two instances of pressing $\boxed{+10}$ that have their value increased 2 times

*Two instances of pressing $\boxed{+10}$ that have their value neither increased or decreased

*One instances of pressing $\boxed{-10}$ that have their value decreased 1 time


Therefore in total you'll need to press $\boxed{+10}$ four times, and $\boxed{-10}$ one time. Somewhere between them you'll need to press $\boxed{\times 10}$ and $\boxed{÷10}$ appropriate number of times, so that two of the $\boxed{+10}$ had their value increased twice, two $\boxed{+10}$ had their value unchanged, and $\boxed{-10}$ had their value decreased once.
There are actually three ways to do this in a minimal number of operations. One is the one you've found:
$$\boxed{+10}\boxed{+10}\boxed{\times 10}\boxed{\times 10}\boxed{+10}\boxed{+10}\boxed{\times 10}\boxed{-10}\boxed{÷10} $$
The other two are
$$\boxed{+10}\boxed{+10}\boxed{\times 10}\boxed{\times 10}\boxed{+10}\boxed{\times 10}\boxed{-10}\boxed{÷10}\boxed{+10} $$
$$\boxed{+10}\boxed{+10}\boxed{\times 10}\boxed{\times 10}\boxed{\times 10}\boxed{-10}\boxed{÷10}\boxed{+10}\boxed{+10} $$
