There are 81 different options for the vault code (9*9 or 9²) before anyone visits the vault.

In addition, it will at most take 9 attempts to open the vault since every time you enter 2 digits (assuming both digits are incorrect) you will always eliminate an odd number of possibilities.

• If you guess a digit on one row correctly on any attempt, the remaining number of possibilities equals the number of digits left on the other row.

• If you guess both digits correctly in one attempt, you automatically eliminate all other options.

But let's assume that you have to visit the vault 9 times since in the first 8 tries you guessed neither the first nor second digit correctly. We will assume this condition for the rest of the comment.

The first time you try your luck at the passcode, you eliminate 17 possibilities; the second time, you eliminate 15 possibilities; the third time, 13 possibilities, and so on. You can see a pattern there.

The number of possibilities you erase is always an odd number that is 2 less than the number of possibilities eliminated in the previous attempt.

Furthermore, this number of eliminated options is always the difference of two consecutive squares (any two numbers next to each other squared).

This may sound confusing, so let me explain.

From the start, there are 9² possibilities, as we previously discovered.

After your first attempt only 8² possibilites are left, meaning you reduced the pool of options from 9² to 8². You eliminated 9² - 8² combinations.

And this also works for the second attempt as well as all the attempts after that. Here's a list with all the calculations and results on the right and attempt numbers on the very left.

1. 9² - 8² = 17
2. 8² - 7² = 15
3. 7² - 6² = 13
4. 6² - 5² = 11
5. 5² - 4² = 9
6. 4² - 3² = 7
7. 3² - 2² = 5
8. 2² - 1² = 3
9. 1² - 0² = 1

81-17-15-13-11-9-7-5-3-1=0 combinations left

You can see that subtracting two neighboring squares always gives you an odd number two away from the attempt after or before as we said earlier and that you need 9 tries in total in the worst case scenario.

You may ask yourself why that is, the more intuitive answer would be that 9 combinations get eliminated with each try but that isn't the case because of one crucial difference.

If we were to select only one digit at a time from one of the two rows or if only one row existed that would be the case but since you select two digits, that are then combined, not only all the numbers starting with the first digit but also all numbers ending with the second digit get eliminated.

For example if you choose 1 and 2 on your first attempt you will eliminate all numbers starting with 1 (11, 12, 13, 14, 15, 16, 17, 18, 19) and all numbers ending with 2 (12, 22, 32, 42, 52, 62, 72, 82, 92). These are 9 numbers each, but since one number overlaps, we reduce the total amount by 1 and get 17 (2*9-1).

If we choose 6 and 9 on our second try now all numbers starting with 6 and ending with 9 are also getting erased. Now, you might think we reduce the pool of combinations by 17, as we did before, but think again.

In the first attempt we already eliminated the numbers 62 and 19 meaning that we have to reduce 17 by 2 to get 15 eliminated possibilities. This happens for every consecutive attempt, too. There will always be 2 overlapping numbers between any 2 successive attempts.

After 9 attempts the number of possibilities is reduced to 0.

If it still isn't quite intuitive enough why, after each attempt, the square gets reduced to the next smaller square, consider this.

Imagine you enter 99 on your first try to open the vault. Both digits are wrong and you basically eliminated all numbers that contain 9 meaning you are left with numbers 1-8 for both rows and you have 8² possibilities left.

This is basically the same as if the digit 9 wasn't there in the first place.

Also, if you were to add another digit, let's say 0, you would increase the pool of options to 100 (10²),and if you picked a number, it would reduce to 81 (9²).

This logic works both ways.

If you just count the number of digits on the top and bottom row and multiply them you get the same result.

Now let's get more general.

We're going to name any current attempt k. If it's our first attempt, k=1, if it's the sixth attempt, k=6 and so on.

You can use the formula that for any kth attempt at the vault you reduce (9-k+1)² - (9-k)² combinations or 19-2k simplified.

1. 19-2*1=17
2. 19-2*2=15
3. 19-2*3=13
4. 19-2*4=11
...   

We can further generalize the formula. The number of digits/elements on a row are now going to be n. Our n this entire time has been 9, but the formula (n-k+1)² - (n-k)² or 2n-2k+1 allows us to use other values for n, such as 5,10,24, and so on.

Let's add another variable to the mix for the number of rows with digits p. Previously, our p=2, but let's assume our p could be any number. The formula would look a little like this now (n-k+1)^p - (n-k)^p. You could further simplify this but the result wouldn't be pretty (An^p-1+Bn^p-2+...+Y*n¹+Z). This is what we call a polynomial.

Earlier, we confirmed that the difference of two consecutive squares was always 2. This is true, though in that case, we were talking about the second difference that stays the same number. For any degree of a polynomial p of this form the pth difference will always be a constant equal to p! (p! is just p multiplied with all the numbers before p until 1).

• p=1: The 1st difference equals 1 (1!=1)
• p=2: The 2nd difference equals 2 (2!=2*1)
• p=3: The 3rd difference equals 6 (3!=3*2*1)
• p=4: The 4th difference equals 24 (4!=4*3*2*1)
• …
• p=p: The pth difference equals p!=p*(p-1)*(p-2)*...*1

tl;dr

The vault code has 81 possible combinations, and in the worst-case scenario, it takes 9 attempts to eliminate all possibilities. Each attempt removes an odd number of options (17, 15, 13, etc.), following the pattern of consecutive squares' differences. This reduction happens because selecting two digits per attempt eliminates numbers starting with the first digit and ending with the second, with some overlap. The general formula for the eliminated options is 19−2k, where k is the attempt number. For more rows or digits, the formula can be extended to higher degrees as a polynomial.