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Question:
Arrange the digits 1,2,3,4,5,6,7,8,9,0 to make a ten-digit Number that satisfies all of the divisibility rules for 2,3,4,5,6,8,9,10,&11. BONUS: make the number also divisible by 7

Alright. First of all,

3 and 9 are automatically okay. The sum of the digits from 0 to 9 is 45, which is divisible by 3 and 9, so any number made out of these ten will be divisible by 3 and 9.

Now,

the last digit MUST be 0, to satisfy 2 and 5 and 10. This also satisfies 6 as well, because it's now divisible by 2 and 3.

Now, let's look at the second-last and third-last digits.

The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.

Now

all we have is 11 (and 7 for extra credit). To satisfy this, we have to make it so that when placing alternating signs between the digits so the result is a multiple of 11. After a bit of trial and error, I came up with 4123975680, which works for everything. It's not divisible by 7, though. I'll keep looking.

This is a good problem to attack by computer:

It turns out that there are 7344 solutions, including 1056 to the bonus.

So I thought, how far can we go?

And it just happens that 2438195760, 3785942160, 4753869120 and 4876391520 are exactly those permutations of 0123456789 divisible by each of 1 through 18, and none of them are divisible by 19.

Code used (Python 3 IDLE):

>>> import itertools
>>> normal=
>>> bonus=
>>> extra=
>>> for i in itertools.permutations('123456789'):
... n=int(''.join(i+('0',)))
... if 0==n%8==n%9==n%11:
... normal.append(n)
... if 0==n%7:
... bonus.append(n)
... if 0==n%13==n%16==n%17:
... extra.append(n)
...
>>> len(normal)
7344
>>> len(bonus)
1056
>>> extra
[2438195760, 3785942160, 4753869120, 4876391520]
>>> [i%19 for i in extra]
[12, 13, 17, 5]