An Autobiography in Sixteen Characters

Before we start the problem, let’s make sixteen blanks. Let’s also slice up this number into five different categories, like this:

Next, let’s get some key insights on this number. One thing we know for sure is the sum of the digits, which should be $16$ (this is in decimal notation, in hexadecimal, it would be $10$ ). This is because each digit shows how many of a specific digit exists in the number. When accounting all of the digits, we should indeed get that the digit sum should equal the length of the number. This is important to realize, so if you don’t get it, look at Brilliant’s quiz on autobiographical numbers to help you.

Now, let’s look at Category $V$ . Can all of these be zero? No, because then we would have at least eight zeros, which means the first digit would be at least eight. This means that there must be a non zero digit in Category $V$ , a contradiction to our initial statement. Moreover, we can’t have a digit sum greater than two in Category $V$ , because the digit sum of the entire numbers will be greater than sixteen! Thus, there must be all zeros except for one one in Category $V$ .

That pesky $1$ means that there must be a $8$ to $F$ digit somewhere. Since we already have seven zeros, the $8$ to $F$ digit must be the first digit.

In addition, the second digit cannot be zero because we have a one in Category $V$ , and it cannot be one, because then there would be at least two ones (one in Category $V$ and one in Category $II$ ). This means that the second digit must be at least two.

With that, we have at least $11$ of the $16$ digit sum counted for. Now, let’s turn to Category $IV$ . If there was just one one in this category, even if it was in the place that counts fives, we would exceed our digit sum of $16$ , because $8+2+1+(5+1)=17$ . Try to see why that is true yourself. Thus, all digits in Category $IV$ are zeros.

Phew! This is what we have so far:

Now, for Category $III$ , if we only consider the digit sum, putting a one in this category shouldn't cause a problem, but putting two or more will exceed sixteen, so we only have to consider placing ones. A little bit of effort shows that if you place a one in any of these spots, there won’t be enough space to fit the resulting digits. Thus, all blanks in this category must be zeros, too.

With that, this is where we are:

It is obvious now that the sixteen digit hexadecimal autobiographical number is $C210000000001000$ . There are $\boxed{12}$ zeros in this number.

According to this article on self-descriptive (or autobiographical) numbers: "In bases $7$ and above, there is, if nothing else, a self-descriptive number of the form ${(b-4)b^{b-1}+2b^{b-2}+b^{b-3}+b^{3}}$ , which has $b-4$ instances of the digit $0$ , two instances of the digit $1$ , one instance of the digit $2$ , one instance of digit $b - 4$ , and no instances of any other digits."

In hexadecimal, $b = 16$ , so there are $b - 4 = 16 - 4 = \boxed{12}$ instances of the digit $0$ . (The actual hexadecimal autobiographical number is $C210000000001000$ .)

I just intuitively guessed the first digit to be F with rest being zeroes. It took me about 4-5 iterations of fixing until reaching the C21000...1000 autobiographical number. Luck i guess 😗