**Part 1: Prove that there cannot exist a sequence of 14 consecutive special numbers:**
Consider a special number `x` that ends in 3 (that is, 3 is the least-significant digit).
`x` must be divisible by three, because it's special and contains a 3 as one of its nonzero digits.
As with any number `x` divisible by three, `x+3, x+6, x+9, ...` are all also divisible by
three, and so are `x-3, x-6, x-9, ...
`

`x+10` and `x-10` cannot be divisible by 3, but they also contain a 3 (the last digit),
and are therefore not special. In other words, a series of 10 or more consecutive special numbers can contain
at most one that ends in 3.

The same argument applies to `x` ending in 4,6,7,8, and 9 (but not to 0, 1 and 2).
So a series of special numbers can end in 0,1,2,3,4,5,6,7,8,9,0,1,2 (thirteen numbers), but fourteen
is impossible, because such a series would have to contain `x` ending in 3
(or 4, or 6, or 7,8, or 9) and also `x+10`.

**Part 2: Find a sequence of 13 consecutive special numbers:**
The obvious answer is
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
except that zero is not special (the puzzle defines zero as nonspecial).

We're looking for a sequence of thirteen numbers that end in the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2 such that all are special.

Let's name the first number in the list `X`.
We know that
`X` is a special number that ends in zero, and
`X+1`, `X+2`, ..., `X+12` are all
special, too.

Now, `X` is some (possibly large) number, and it ends in 0.
It cannot
contain the digit 2. If it did, `X+1` would also contain
2, but `X+1` ends in 1, and is therefore odd and not divisible
by 2 (and therefore not special). In fact, following the same
reasoning, it cannot contain *any* digit 2-9. In can
however, contain only the digits zero and one.
In addition to this, it must not only end in a zero, but actually
end in *two* zeros, because if the second-to-last digit in `X`
was a 1, then the second-to-last digit in `X+11` would be 2,
and `X+11` would be an odd number that contains a 2, and
therefore not special.

Following similar reasoning, `X` must be divisible
by two. This is because `X+2` will end in a 2, and
since `X+2` must be special, it must be divisible by 2,
and so that number minus 2 must also be divisible by 2. This
reasoning applies to all digits. `X` must be divisible
by 7, because `X+7` must be divisible by 7 (since it
must be special and it will contain a 7 as the last digit).
So, `X` must contain only zero and ones, end in two
zeros, *and* be divisible by all digits. (You'll see
that we get `X` plus 10, 11, and 12 for "free", because `X` plus
ten and eleven result in numbers that contain only zeros and
ones, all of which are special, and `X+12` contains only
zeros and ones except for the final number 2, which doesn't hurt;
all numbers ending in 2 are divisible by 2).

Now all we need is a number, `X`, containing only zeros and ones,
ending in *two* zeros and divisible by all digits.
No problem.

- A number is divisible by 2 if the number ends in 0 or an even number.
- A number is divisible by 3 if the sum of the digits is divisible by three.
- A number is divisible by 4 if the number can be divided by two twice.
- A number is divisible by 5 if the number ends in a zero or a five.
- A number is divisible by 6 if the number is divisible by both 2 and 3.
- A number is divisible by 7 if it is. We'll have to actually do the division.
- A number is divisible by 8 if the number can be divided by two thrice.
- A number is divisible by 9 if the sum of the digits is divisible by nine.

There are ways to "construct" a number that is divisible by every digit by
supplying it with the properties above, although the digit 7 is more difficult.
Let's define another number, `S`, which contains only zeros and ones.
If the number of ones sums to a multiple of nine, `S` is divisible by
nine. If it doesn't, I can multiply `S` by 10^{N} (where
`N` is the number of digits in `S`) and add `S`.
This produces a new number, `S'`, which is written as `SS`
(that is, the sequence of digits in `S` written twice). `S'`
is still divisible by `S` (and therefore everything `S`
was divisible by), because only integer multiplication and adding `S`
was applied. This operation can be repeated (it will take at most nine such
operations), to
produce a number whose digits must sum to a multiple of nine. Now, I multiply
the resulting number by 1000 (this ensures that it ends in two zeros).
It is now divisible by all digits (except maybe 7):

- It's divisible by 2 because it's even.
- It's divisible by 3 because it's divisible by 9.
- It's divisible by 4 because it's divisible by 1000, which is divisible by 4.
- It's divisible by 5 because it ends in zero.
- It's divisible by 6 because it's divisible by 2 and 3.
- It's divisible by 8 because it's divisible by 1000, which is divisible by 8.
- It's divisible by 9 because we made sure of that.

If the original number `S` happens to be divisible by 7,
the result of these operations will too, so it would be good to start
with a value for `S` which is divisible by 7. The smallest
number which qualifies is 1001. We can find the number 1001
with a little guess-and-check (it's
only the ninth positive integer that contains only zeros and ones).
Using 1001, and applying the operations described, results
in the number
`X` as
100110011001100110011001100110011001000.
The sequence

100110011001100110011001100110011001000 100110011001100110011001100110011001001 100110011001100110011001100110011001002 100110011001100110011001100110011001003 100110011001100110011001100110011001004 100110011001100110011001100110011001005 100110011001100110011001100110011001006 100110011001100110011001100110011001007 100110011001100110011001100110011001008 100110011001100110011001100110011001009 100110011001100110011001100110011001010 100110011001100110011001100110011001011 100110011001100110011001100110011001012is a sequence of 13 consecutive special numbers. Of course, there are sequences of 13 consecutive numbers that contain smaller numbers. (Okay, perhaps it is interesting to find the smallest, but this puzzle doesn't ask for that). It would be nice to have a more direct method for finding an integer containing only zeros and ones that is divisible by 7 (although I doubt there is a simple one).