Can you solve the Leonardo da Vinci riddle? - Tanya Khovanova
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.
