66336:
The last digit is 6 because 3 is not divisible by 2.
The second-to-last digit is 3 because 66 is not divisible by 4.
The third-to-last digit is 3 because 636 is not divisible by 8.
The fourth-to-last digit is 6 because 3336 is not divisible by 16.
The fifth-to-last digit is 6 because 36336 is not divisible by 32.
(Edit: This is sufficient because $2^{n}$ divides $10^{n}$, so it doesn't matter what the preceding digits are.)
All numbers will be of the form $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$ where $a_i+b_i=9$ for all $i$ using each digit exactly once and each number of that form will satisfy your conditions. Proof below.
As such, by choosing $a_1$, the choice of $b_1$ is forced. Similarly choosing $a_2,a_3,a_4,a_5$ will force the choice for $b_2,b_3,b_4,b_5$.
Applying multiplication principle, and remembering that leading zeroes do not contribute to the number of digits of a number (e.g. $012349876$ does not count as a ten-digit number) we have $9$ options for $a_1$, $8$ options for $a_2$, $6$ options for $a_3$, $4$ options for $a_4$, and $2$ options for $a_5$.
There are then $9\cdot 8\cdot 6\cdot 4\cdot 2=3456$ ten digit numbers satisfying all of the desired properties.
The largest number of which is formed with the largest selections available for $a_1,a_2,\dots$ respectively and is then $9876501234$, the $10000$'s place being the $5$.
Lemma: Any ten digit number of the form $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$ is divisible by $11111$ if and only if $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5$ is divisible by $11111$.
$a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5\equiv (9\cdot 11111)a_1a_2a_3a_4a_5+a_1a_2a_3a_4a_5 + b_1b_2b_3b_4b_5$
$\equiv 10^5a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5\equiv a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5\pmod{11111}$
Claim: In the set of ten digits numbers using all digits $0$-$9$ the number is divisible by $11111$ if and only if it is of the form $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$ where $a_i+b_i=9$ for all $i$.
$\Leftarrow)~~$ Suppose that $a_i+b_i=9$ for all $i$. Then $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5=99999$ is divisible by $11111$. Then by the lemma, so too is $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$.
$\Rightarrow)~~$ Suppose that $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$ is a ten-digit number using all digits exactly once which is divisible by $11111$.
By the lemma, we know then that $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5=n\cdot 11111$ for some $n$.
As all digits are used exactly once, the sum of the digits is $1+2+\dots+9=\frac{9\cdot 10}{2}=45$. As this is divisible by $9$ both $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$ and $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5$ must also be divisible by $9$. (see sum of digits of a number is equivalent modulo 9 to the number itself proof elsewhere).
As the digits are distinct $16047= 13579+02468\leq a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5\leq 97531+86420=183951$ as those selected numbers minimize and maximize the sums of the respective digit positions respectively. From this, we know then that $1\cdot 11111<a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5<17\cdot 11111$ so we know that $n\in \{2,3,\dots,16\}$.
As $11111$ is coprime to $9$, by chinese remainder theorem, we know then that $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5$ must be divisible by $9\cdot 11111=99999$. The only value then for $n$ which satisfies this is $n=9$ and therefore $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5=99999$
Now... if $a_i+b_i\neq 9$ for some $i$ then there must be some $a_j+b_j>9$ for some $j$ due to law of averages. The largest value of which is $9+8=17$, and the largest carryover for a specific digit place is $1$, which would result in a digit of $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5$ being strictly less than $9$. It follows then that it must be that $a_i+b_i=9$ for all $i$.
Best Answer
Let $n=100a+10b+c,$ where $a> 0$ and $b,c\geq 0$. We are trying to solve $$100a+10b+c=a+b+c+ab+ac+bc+abc \\ \implies 99a+9b=abc+ab+ac+bc \\ \implies a(99-b-c-bc)=b(c-9) \\$$$c-9\leq 0$, but $b+c+bc\leq 99$. So the above equation holds iff $b=c=9$, which means $a$ can take any value.