Estimating the minimum number of the digit from above by a linear function

Estimating the minimum number of the digit from above by a linear function

For any natural number $k$, there are natural numbers $n$ which satisfy
the following three conditions.
1. Each digit of n is either $1$ or $0$.
2. There are $k$ occurrences of the digit $1$.
3. $n$ is a multiple of $k$.
One of the proofs is the following.
Proof: Considering the remainders when you divide each of
$1,10,10^2,\cdots$ by $k$, at least one of the numbers from $0$ to
$k−1$ appears infinitely as a remainder. Let it be $r$ and suppose
$10^{a_1}, 10^{a_2}, \cdots, 10^{a_k}$ as the numbers whose remainders are
$r$ divided by $k$. Note that $a_1, a_2, \cdots, a_k$ are non-negative
integers and that they are different from each other. Then,
$n=10^{a_1}+10^{a_2}+\cdots+10^{a_k}$ satisfies the three conditions, so
the proof is completed.
In fact, there are a infinite number of $n$ for a given $k$.
Now, let the number of digits in the smallest such $n$ for a given $k$ be
$d_k$.
For $$k=1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,$$ $d_k$ is
$$1,3,3,6,6,7,8,11,9,11,21,14,15,16,16,20,18,19,20,22$$ respectively.
Obviously, $d_k\ge k$ for any $k$. I've already got the following three.
1. $d_k=k$ only if $k=1, 3, 9$.
Proof: Let $S(n)$ be the digit sum for a natural number $n$, and difine
Proposition 1 as the following.
Proposition 1: If $S(m)$ is a multiple of $n$ for a natural number $m$,
then $m$ is a multiple of $n$.
It is sufficient to prove the following.
"Proposition 1 is true only if $n=1, 3, 9$".
Let's prove this. Suppose that proposition 1 is true. Then, for
$$m_1=10^{n}+10^{n-1}+\cdots+10^2+1, m_2=10^{n}+10^{n-1}+\cdots+10^2+10,$$
$S(m_1)=S(m_2)=n$, so each of two is a multiple of $n$. This leads that
$m_2-m_1=9$ is a multiple of $n$ too. Hence, $n$ is a divisor of $9$.
Then, the proof is completed.
2. For any $k$, $d_k\le k(k−1)+1$.
Proof: If the digit of $n$ is $k(k-1)$, then there is some possibility of
every remainder appearing $k-1$ times. If $k(k-1)+1$, then there is no
possibility of that.
3. If $10$ is the primitive root of mod a given prime $p$, then $d_p=p+1$.
Proof: By Fermat's little theorem, $10^{p-1}\equiv1$(mod $p$). Noting that
$$1+10+\cdots+10^{p-2}\equiv1+2+\cdots+(p-1)=\frac{p(p-1)}{2}\equiv0,$$
$p$-digit is impossible because of the following.
$$1+10+\cdots+10^{p-1}\equiv10^p+\left(1+10+\cdots+10^{p-2}\right)\equiv10^p
\not\equiv 0$$
In addition to this, noting that $10$ is not the primitive root of $11$,
we can say $$1+10+\cdots+10^{p}\equiv10^{p+1}+10^p\not\equiv0.$$Hence,
there is $r\ (1\le r\le {p-1})$ such that $$10^{p+1}+10^p\equiv10^r$$By
changing $1$ to $0$ at the digit in the $10^r$'s place, we can make $n$
which satisfies the three conditions. Now the proof is completed.
Then, here is my questions.
Question1: Estimate $d_k$ from above by a linear function of $k$.
For example, can we get something like $d_k\le 2k$?
Question2: What $k$ does give the maximum of $d_k-k$?
For example, $d_{11}=101010101010101010101$ and $d_{11}-11=10$. Does $11$
give the max?
I'm facing difficulty. I need your help.