Finding the least 100-digit number, which has no $0$ in its decimal representation, that is divisible by the sum of its digits. An interesting question on number theory:

Find the least 100-digit number, which has no $0$ in its decimal representation, that is divisible by the sum of its digits.

My attempt: let the number be
$$a_1 a_2 a_3.....a_{100}$$
Which can also be written as
$$10^{99}a_1 + 10^{98}a_2+\cdots +10a_{99}+a_{100}=(10^{99}-1)a_1 + (10^{98}-1)a_2+\cdots +(10-1)a_{99}+(a_1 +a_2 + ...a_{99} + a_{100})$$
The last term is divisible by the digit sum. But I don't know how to proceed further.
Thanks for any help.

A slightly better answer than "by computer" but still requires trial and error: this uses Fermat's Little Theorem.
Obviously ..$11$, ..$13$, ..$15$, ..$17$, ..$19$ are not divisible by their digit sums, which are all even.
..$16$ is not divisible by its digit sum $105$.
Since ..$11$ is divisible by $101$, neither ..$12$ nor ..$21$ are divisible by their digit sum $101$.
The harder part is to check ..$14 \pmod {103}$, ..$18 \pmod {107}$ and $..22 \pmod {102}$.
Note that ..$14 = $..$11 + 3 = (10^{100}-1)/9 + 3 = (10^{100}+26)/9$. Since $103$ is prime, finding $10^{100} \pmod {103}$ would help find ..$14 \pmod {103}$.
Now by FLT, $10^{102} \equiv 1 \pmod {103}$. Also by using Euclidean Algorithm or otherwise, we have:
$103 \times 3 = 309 = 10\times 31-1$, so $10^{-1} \equiv 31 \pmod{103}$.
Finally ..$14\times 9 = 10^{100}+26 \equiv 10^{-2} + 26\equiv 31^2 + 26\equiv 60\pmod {103}$, so ..$14$ is not divisible by $103$.
I leave ..$18 \pmod {107}$ to you.
For $..22 \pmod {102}$, since $102 = 2 \times 3 \times 17$ and divisibility by $2$ and $3$ are very easy to check, we just need to find $(10^{100}-1)/9 + 11 \pmod {17}$, and showing $10^{100}+98\equiv 10^4 + 98$ is divisible by $17$ is trivial.
