Generalize multiples of $999...9$ using digits $(0,1,2)$ The smallest $n$ such that $9n$ uses only the three digits $(0,1,2)$ is $1358$, giving a product $12222$. For $99n$ this is $11335578$, giving $1122222222$. Similarly, 
$999(111333555778)=111222222222222,$
$9999(1111333355557778) = 11112222222222222222, ...$ 
The product seems to be $k$ 1's followed $4k$ 2's for $k$ digit $999...9$. For $n$, this seems to be $1..3...5...7...8$, with $1, 3, 5$ $k$ times and $7$ $k-1$ or $0$ times. Can this pattern be proven and extended for any $k$? 
 A: I think this helps:
$9n$ is a multiple of $9$ so its sum of digits $\equiv0 \mod 9$
Consider :


*

*number of $1$s in $9n$ is $x$

*number of $2$s in $9n$ is $y$

*number of $0$s in $9n$ is $z$


$$x + 2 y + 0z \equiv 0 \mod 9$$
Solve this when number of digits $(x + y + z)$ is minimum (this means when $9n$ is minimum)
It happens when $z = 0$
Solve $x + 2 y \equiv 0 \mod 9$
When $x + y$ is minimum
This happens when $y = 4x$
If you find $x$ and $y$, minimum of $9n$ is in the form $111111...2222...$
$1$ appears $x$ times and $2$ $y$ times in the number
You only need to show :
$99... (9 \text{ appears } x \text{ times} ) \times  11...33...555..77..8  (1 , 3 , 5~x \text{ times}, 8)$
equals $11...222... ( 1 : x \text{ times},~2 : 4x \text{ times} )$
A: It is readily verified that 
$$\tag1(10^k-1)\cdot\frac{10^{4k}+2\cdot 10^{3k}+2\cdot 10^{2k}+2\cdot 10^k+2}{9} =\frac{10^{5k}-1}9+\frac{10^{4k}-1}{9}$$
and that the two fractions are in fact integers and the number on the right, being the sum of two repunits, is written with $1$ and $2$ only.
Remains to show that the numbers in $(1)$ are minimal.
Note that $(10^k-1)n=10^kn-n$ is the difference of two numbers of different length, so that $10^kn$ (as well as $n$) must start with at least $k-1$ digits $\in\{0,1,2\}$ and the $k$th digit $\in\{0,1,2,3\}$; actually, a $3$ as $k$th digit is only possible if it is followed by a digit $\le 2$ as otherwise no carray (or rather: borrow) orccurs. Also, $(10^k-1)n\equiv -n\pmod{10^k}$ so that the last $k$ digits of $n-1$ must be $\in\{7,8,9\}$. Note that 
$$\left\lfloor\frac{(10^k-1)n}{10^k}\right\rfloor=n-\left\lceil\frac{n}{10^k} \right\rceil=(n-1)-\left\lfloor\frac{n}{10^k} \right\rfloor$$
so that we conclude about the next $k$ digits: We subtract someting from digits $7,8,9$ and obtain digits $0,1,2$; this is only possible if no  borrow occurs, hence the next $k$ digits of $n-1$ are $\in\{5,6,7,8,9\}$. By repeating the argument, the next $k$ digits of $n-1$ are in $\{3,4,5,6,7,8,9\}$. Therefore, $n$ either has more than $4k$ digits, or $n-1$ is described by the regular expression [12][012]{k-1}[3-9]{k}[5-9]{k}[7-9]{k}. We can exclude the case of more than $4k$ digits as $(1)$ already beats that.
Thus we are left with the second case that can also be formulated as:
$$ n=\alpha\cdot 4^{3k}+\beta\cdot10^{2k}+\gamma\cdot 10^k+\delta+1$$
where $\alpha,\beta,\gamma,\delta$ are $k$-digit number with digits $\in\{0,1,2\}$, $\in\{3,\ldots,9\}$, $\in\{5,\ldots,9\}$, $\in\{7,\ldots,9\}$, respectively. Then 
$$(10^k-1)n=\alpha\cdot 10^{4k}+(\beta-\alpha)\cdot 10^{3k}+(\gamma-\beta)\cdot 10^{2k} +(\delta-\gamma)\cdot10^k+(10^k-1-\delta)$$
A stated above, the subtractions $\delta-\gamma,\gamma-\beta,\beta-\alpha$ involve no borrows. Therefore any zero occuring in $\alpha$ would produce a digit $\ge 3$ in $\beta-\alpha$ and hence also in $(10^k-1)n$. We conclude that $\alpha$ contains no zeroes. But then $\alpha \ge 11\ldots 1$, $\beta\ge 33\ldots 3$, $\gamma\ge 55\ldots 5$, $\delta\ge 77\ldots 7$ and ultimately the solution $(1)$ is inded the minimal solution.
