Riddle: A special $6$-digit number Here is a riddle:

Riddle: I am thinking about a $6$-digit number
  $
\underline{ }\,
\underline{ }\,
\underline{ }\,
\underline{ }\,
\underline{ }\,
\underline{ }
$
  (no leading zeros). All digits are distinct. When I multiply the number by $2$, I get another $6$-digit number with the same digits but mixed up. The same happends when multiplying by $3,4,5$ and $6$. What number I am thinking about?

I know the solution, as well as a way to find it (see below). But my approach uses the computer for some annoyingly exhausting multiplication tasks. I want to demonstrate an approach on a whiteboard without help of too much computing power (calculator is okay of some few computations) and too deep mathematics.
Question: Can you think of a nice/elegant human-only way to approach the solution?

Spoiler!!
You were warned. It is fun to think about the solution yourself. But here it is:

 $142857$.

So here is my approach. Lets denote by $x\in\Bbb N$ the number we are looking for. Because $6x$ is still a six-digit number, we know $6x\le 999999\Rightarrow x\le166666$. This gives us the first digit:
$$
\underline{1}\,
\underline{ }\,
\underline{ }\,
\underline{ }\,
\underline{ }\,
\underline{ }
$$
Now we can see that the first (left-most) digit of $nx$ is always different for different values of $n$, but never $0$. As this generates all the six possible digit of $x$, we see that zero cannot be on of those. Hence, we can exclude it in the following explanation.
Next, we are trying to find the last (right-most) digit. It cannot be $1$ because the digits are distinct. So the remaining options are $2,3,4,5,6,7,8,9$. This last digit alone determines the last digit of the multiples $nx,n=1,...,6$. Here is how:
$$
\begin{array}{c|cccccccc}
\times & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
1 &  2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
2 &  4 & 6 & 8 & 0 & 2 & 4 & 6 & 8 \\
3 &  6 & 9 & 2 & 5 & 8 & 1 & 4 & 7 \\
4 &  8 & 2 & 6 & 0 & 4 & 8 & 2 & 6 \\
5 &  0 & 5 & 0 & 5 & 0 & 5 & 0 & 5 \\
6 &  2 & 8 & 4 & 0 & 6 & 2 & 8 & 4 \\
\end{array}
$$
The number in row $i$ and column $j$ shows the last digit of $ix$ when the last digit of $x$ is $j$ (computed via $ij\text{ mod }10$). We can exclude $2,4,5,6$ and $8$ because the last digit cannot become $0$. The remaining digits $3,7$ and $9$ all generate distinct last digit for different factors, but only $7$ generates the necessary digit $1$. Hence we see that the last digit must be $7$:
$$
\underline{1}\,
\underline{ }\,
\underline{ }\,
\underline{ }\,
\underline{ }\,
\underline{7}
$$
This also yields all digits of $x$, namely $1,2,4,5,7,8$ and we just have to arrange them appropriately. We can exclude $8$ for the second digit (from the left) and with some more elaboration we can also exclude $5$. But the main part of my work now becoms computationally exhausting in nature: we now look at the last two digit and plug in all possible values to obtain $27,47,57,87$. These completely determine the last two digits of $nx,n=1,...,6$. We then write down another table like the one above and exclude all but one combination for different (but straight forwards) reasons. This gives
$$
\underline{1}\,
\underline{ }\,
\underline{ }\,
\underline{ }\,
\underline{5}\,
\underline{7}
$$
Now we do this for the last three digits, then the last four $-$ and so on $-$ until we are left with the solution mentioned above. Notably, the solution is unique.
As you can see, the generation of these multiplication tables is pretty annoying and will bore anyone who is listening to me. I would really welcome a more elegant approach.

Of course, given the solution it is very easy to check that it indeed is one. But the elegance I am looking for is a way to see this without actually computing $nx,n=1,...,6$. Here are some observations on the solution that might help in finding a more elegant approach.


*

*The solution is exactly $999999\div 7$.

*The solution is one period of the deciaml expansion of $1/7$.

*The digit permutations are actually just cyclic shifts.


What is so special about $7$ that it can be used to generate such an amazing number? Maybe the way to go is looking for generalizations.
 A: You already know that the number has to be of the form $$
\underline{1}\,
\underline{ }\,
\underline{ }\,
\underline{ }\,
\underline{ }\,
\underline{\color{green}{7}}
$$
Now since
$$\color{green}7\times 2=1\color{red}{4},\quad \color{green}7\times 4=2\color{red}{8}, \quad \color{green}7\times 5=3\color{red}{5},\quad \color{green}7\times 6=4\color{red}{2}$$
we see that the other digits have to be $4,8,5,2$.
The last two digits have to be $57$ since
$$27\times 4=1\color{red}{0}8,\quad 87\times 3=2\color{red}{6}1,\quad 47\times 2=\color{red}{9}4$$
The last three digits have to be $857$ since
$$457\times 2=\color{red}{9}14,\quad 257\times 4=1\color{red}{0}28$$
The last four digits have to be $2857$ since
$$4857\times 2=\color{red}{9}714$$
A: Starting with 
$$x=1 \,\_\,\_\,\_\,\_\, 7$$
We also know, that the rest of digits are $4,8,5,2$ (from your table)
Now, as we double our number we obtain
$$2x=2 \,\_\,\_\,\_\,\_\, 4 \\\text{ or }\\2x= \color{red}{3} \,\_\,\_\,\_\,\_\, 4 $$
$3$ is not in our set of digits, thus
$$2x=2 \,\_\,\_\,\_\,\_\, 4$$
To obtain $2$ as first digit in $2x$, the second digit in $x$ have to be less than $5$, so
$$x=12\,\_\,\_\,\_\, 7 \\\text{ or }\\x= 14\,\_\,\_\,\_\, 7 $$
Let's consider 2 cases:


*

*$$x=14\,\_\,\_\,\_\, 7$$
In this case 
$$2x=28\,\_\,\_\,\_\, 4$$
To obtain $8$ as second digint in $2x$, the third digit in $x$ have to be less than $5$, so
$$x=142\,\_\,\_\, 7$$
and
$$2x=285\,\_\,\_\, 4$$
$5$ as the third digit in $2x$ means, that $4$th digit in $x$ is greater or equal $5$. Both of our remaining digits are $\geq 5$, so let's try both of them:
$$x_1=142857\text{ or } x_2=142587$$
multiplying this we obtain:
$$2x_1=285714\text{ or } 2x_2=285174$$
$$3x_1=428571\text{ or } 3x_2=42\color{red}{776}1$$
$x_2$ multiplied 3 times gave us number with thwo digits $7$ and forbidden digit $6$, so $x_2$ is invalid. Continue multiplying with $x_1$:
$$4x_1=571428$$
$$5x_1=714285$$
$$6x_1=857142$$


Thus  $x=142857$ is proper solution


*$$x=12\,\_\,\_\,\_\, 7 $$
and 
$$2x=25\,\_\,\_\,\_\, 4 $$
$5$ as the second digit in $2x$ means, that $3$rd digin in $x$ have to be $\geq 5$. We have then:
$$x_1=125\,\_\,\_\, 7 \text{ or }x_2=128\,\_\,\_\, 7$$
So:
$$2x_1=251\,\_\,\_\, 4 \text{ or }2x_2=257\,\_\,\_\, 4$$
Repeating this reasoning we obtain:
$$x_1=1258\,\_\, 7 \text{ or }x_2=1285\,\_\, 7$$
$$2x_1=2517\,\_\, 4 \text{ or }2x_2=2571\,\_\, 4$$
In both cases we have to fill the gap with $4$, but it means that $5$th digit in $2x$ would be $9$, but this digit is not allowed, so this case leads to the contradiction.

