No matter what two-digit number you choose, you will always end up at a multiple of nine (9, 18, 27, etc.). The same symbol will be on every multiple of nine so you'll always pick that symbol. Basically, any number minus its own digits will be a multiple of nine.

How does that work, though?

To start, think of the round numbers, 10, 20, 30, etc. Let's start with 10. 10-1=9, because 9 is one less than 10. So it makes sense that two 9s are two less than two 10s, and three 9s are three less than three 10s, and so on. So 20-2=18 (two 9s), and 30-3=27 (three 9s). Essentially x0-x, where x is just a digit, not a number, will equal a multiple of 9.

Okay, but why does it work for all the other numbers too?

The principle is pretty simple. When you add a digit, you also end up taking that digit away when you subtract, so you always end up back at the same number until you reach the next round number (x0), at which point you'll start hitting a new multiple of 9.

If you add one to 10, you get 11. But then because you've added one to the digits, you'll end up subtracting it as well. 11-2=9. Add another one and you get 12, but now the digits equal 3 so you just subtract it off again, 12-3=9. Every time you add one, it'll just get removed. This goes all the way up to 19-10=9, and then at 20 the answer bumps up to 18, another multiple of 9, and the pattern repeats itself with every answer up to 29 being 18 (28-10=18, for example). This works for three-digit numbers as well, and probably for any number with as many digits as you want. The answer will always be a multiple of 9.

As long as they have the same symbol on every multiple of 9, then that's the symbol you're forced to pick.

I'm sure someone here can give a more rigorous explanation of how the math works, but that's the gist as far as I understand it.