Is it possible to solve this problem mathematically? 
On this day, after a total of $4-5$ hours, the following conclusion I reached: This is impossible.

I think there's not a "special" way to do it.I don't even know where to start. I really want to know if at least this is possible or impossible.

 A: I took the puzzle with the constraint that all the 7's are already written on the paper, and that no starred number starts with a zero.

$7\times d$ has only 6 digits, so $d<142858$.

There are two times when d multiplied by a digit has seven numbers, so
$8\times d>1000000$ ($d>125000$). So far, $d=1***7*$.

From the 00*7 that is in the middle, we know that the star must be a 9, and that $7\times d$ starts with $87****$ so the difference (from $97****$) starts with $10****$ and from this, we know that $q=**78*$ and that $d<125714$ ($7\times d<879999$)

Assuming that no other 7 can be written, 8*d has a 7 in second digit, while 9*d has no 7 in it, so $q=*978*$.

$8\times d$ : $10**7**$ and $125070<d<125679$, we can check manually that ($d=125470, d=125471, d=125472, d=125473$ or $d=125474$) are the only solutions that make $8\times d$ in the right form.

With this, only $d=125470$ and $d=125474$ allow $9\times d$ : $1******$ without any other seven. Putting the numbers back in, we quickly reach an impossibility with the "seven minus seven" when we substract $7*d$.

So, I discarded my solution without any extra Seven, and went on the path that Alex Ravsky has just posted - now I am upvoting his solution, without rewriting it.

