How can I prove that starting from $0$, and repeatedly taking $+10,-10,\times10$ or $\div10$, requires at least $9$ operations to get to $2019$?

Solution 1:

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

$$\boxed{+10}\boxed{+10}\boxed{\times 10}\boxed{\times 10}\boxed{+10}\boxed{\times 10}\boxed{-10}\boxed{÷10}\boxed{+10} $$

$$\boxed{+10}\boxed{+10}\boxed{\times 10}\boxed{\times 10}\boxed{\times 10}\boxed{-10}\boxed{÷10}\boxed{+10}\boxed{+10} $$