You’ve found Leonardo Da Vinci’s  secret vault, secured by a series of combination locks. Fortunately, your treasure map  has three codes: 1210, 3211000, and… hmm. The last one appears to be missing. Looks like you’re gonna have  to figure it out on your own. There’s something those  first two numbers have in common: they’re what’s called  autobiographical numbers. This is a special type of number  whose structure describes itself. Each of an autobiographical number’s digits indicates how many times the digit corresponding to that position occurs within the number. The first digit indicates  the quantity of zeroes, the second digit indicates  the number of ones, the third digit the number of twos,  and so on until the end. The last lock takes a 10 digit number, and it just so happens that there’s exactly one ten-digit autobiographical number. What is it? Pause here if you want  to figure it out for yourself! Answer in: 3 Answer in: 2 Answer in: 1 Blindly trying different combinations  would take forever. So let’s analyze the autobiographical  numbers we already have to see what kinds of patterns we can find. By adding all the digits in 1210 together, we get 4 – the total number of digits. This makes sense  since each individual digit tells us the number of times a specific  digit occurs within the total. So the digits in our  ten-digit autobiographical number must add up to ten. This tells us another important thing – the number can’t have  too many large digits. For example, if it included a 6 and a 7, then some digit would  have to appear 6 times, and another digit 7 times– making more than 10 digits. We can conclude  that there can be no more than one digit greater than 5  in the entire sequence. So out of the four digits 6, 7, 8, and 9, only one – if any-- will make the cut. And there will be zeroes in the positions corresponding to the numbers  that aren’t used. So now we know that our number must contain at least three zeroes – which also means that the leading  digit must be 3 or greater. Now, while this first digit counts  the number of zeroes, every digit after it counts how many times a particular non-zero digit occurs. If we add together  all the digits besides the first one – and remember,  zeroes don’t increase the sum – we get a count of how many  non-zero digits appear in the sequence, including that leading digit. For example,  if we try this with the first code, we get 2 plus 1 equals 3 digits. Now, if we subtract one, we have a count of how many non-zero  digits there are after the first digit – two, in our example. Why go through all that? Well, we now know something important: the total quantity of non-zero digits that occur after the first digit is equal to the sum of these digits,  minus one. And how can you get a distribution where the sum is exactly 1 greater than the number of non-zero  positive integers being added together? The only way is for one  of the addends to be a 2, and the rest 1s. How many 1s? Turns out there can only be two – any more would require additional  digits like 3 or 4 to count them. So now we have the leading digit of 3  or greater counting the zeroes, a 2 counting the 1s, and two 1s – one to count the 2s and another to count the leading digit. And speaking of that, it’s time to find out  what the leading digit is. Since we know that the 2  and the double 1s have a sum of 4, we can subtract that from 10 to get 6. Now it’s just a matter  of putting them all in place: 6 zeroes, 2 ones, 1 two, 0 threes, 0 fours, 0 fives, 1 six, 0 sevens, 0 eights, and 0 nines. The safe swings open,  and inside you find... Da Vinci’s long-lost autobiography.