$\square\square\times\square =\square\square\square =\square\times\square\square\,\,\,$ fill blanks with distinct numbers from$\{1,2,3,4,5,6,7,8,9\}$ Fill in the blanks of:

$$\square \;\square \times \square = \square \; \square \;\square =\square \times \square \;\square $$

With distinct numbers from the set $\{1,2,3,4,5,6,7,8,9\}$.
I was able to do it by trial-and-error, but I am looking for a more mathematical approach.
 A: A number of observations can be made that narrow down the number of "guesses" that need to be made:


*

*Neither $x_2$, $x_3$, $x_7$ nor $x_8$ can be a $5$, because this would result in either a $0$ or another $5$ in the evaluation of the product. For the same reason, $x_6$ (the final digit of the product) cannot be $5$.

*Neither $(x_2,x_3)$ nor $(x_7,x_9)$ can be a pair of numbers such that the ones place of their product is one of the multipiers. That is, given a pair of numbers $m$ and $n$, $m\not\equiv mn \pmod{10}$ and $n\not\equiv\pmod{10}$.

*Neither $x_3$ nor $x_7$ can be a $1$, which would result in a $2$ digit product (and a repeat of each digit in the corresponding multiplier).

*You probably want the product to be a multiple of $6.$  It is not guaranteed, but gives more flexibility.

*The only digit pair that multiplies to $1$ is $3\times 7$, so the three-digit number cannot end in $1$. Similarly $3\equiv 7\times 9$ only, $7\equiv 3\times 9$ only, and $9$ is not the product of distinct digits. So the result is even.

A: For general reference, a brute force approach leads to there only being two distinct solutions:
$$29 \cdot 6 = 174 = 58 \cdot 3$$
$$39 \cdot 4 = 156 = 78 \cdot 2$$
Edit:
If you want to see this for yourself:
import itertools
l = itertools.permutations(range(1,10))
for x in l:
    a, b, c = (10*x[0] + x[1]) * x[2], 100*x[3] + 10*x[4]+ x[5], (10*x[6] + x[7])*x[8]
    if (a == b and b == c): 
        print 10*x[0] + x[1], "*", x[2], "=", b, "=", 10*x[6] + x[7], "*", x[8]

A: All right: This is "mathematical". In other words, some reasoning, with a lot of case checking. So, we have that: $$ab\cdot c = de\cdot f = ghi$$


*

*$b,c,f,e,i$ are not 5. Else requires a zero, or another 5.

*If $ghi$ is odd, then we can see that $b,c,e,f$, and $i$ all have to be odd. This is impossible if none of them are a $5$. Therefore, $ghi$ is even.

*$a$ and $d$ are not a $1$. Else, $g$ would also have be a $1$. We can also see that $c$ and $f$ are not $1$. $b$ and $e$ are not $1$ either. If $b$ was $1$, then $i = c$.

*Thus, either $g$ or $h$ is a $1$. 
Now, look at the number $5$. It is either $a,d$, $g$ or $h$.


*

*First, assume that it is $g$. Then $ghi =51i$. And we can see that this is impossible.

*Next, assume it is $h$. then $ghi = 15i$. $i$ is even. So where is the $9$? $9$ cannot be either $a$ or $d$. So assume it is $b$. 
Then we have $a9\cdot c = de\cdot f = 15i$. Since $c$ is odd, $a$ cannot be $2$. If $a =3, c=4$. We have:
$$39\cdot 4 = 156 = 78\cdot 2$$
So if $5= h$, one answer.
Otherwise, $5$ is either $a$ or $d$. Without loss of generality, assume it is $a$, so that
$5b\cdot c = de\cdot f = ghi$
Two cases, if $c$ is odd, $50\cdot c = 250$. We know that $g$ or $h$ is $1$. If $h$ is one, $b*c = 60$ something. That is impossible (only $7*9$ = 60 something, and that is odd).
So if $c$ is odd, $g = 1$. $c$ is then obviously $3$.
So we have $5b \cdot 3 = 1hi = de\cdot f$. b cannot be 2. Else, h is 5. If b is 4, we have 54*3 = 162 = ... impossible, as 7, 8, and 9 are left over.
As for $b =6$, we have $56\cdot 3 = 168$. Not possible. 2 instances of 6.
If b is 8. We have $58\cdot 3 = 174 = 29\cdot 6$
Final situation: if c is even.
$5b \cdot c = de \cdot f = ghi$
if $c$ is 2, then $g$ is 1, and $h$ is either 1 or 0. Impossible.
If $c$ is 4 or more, $g$ is not 1, so $h$ is 1.
We have
\begin{align}
5b \cdot c = g1i = de \cdot f
\end{align}
If $c$ is 4, $b = 3$. Then $53 \cdot 4 = 212$. Impossible.
If $c$ is 6, $g$ is 3. So b is 2. $52 \cdot 6 = 312$. Impossible.
If $c$ is 8, $g$ is 4. b is 2. $52 \cdot 8=416$. No even numbers left. Therefore, impossible.
Basically, that leaves the two answers.
A: Elimination, elimination.
$$\overline{x_1x_2} \cdot x_3 = \overline{x_4x_5x_6}= x_7 \cdot \overline{x_8x_9}$$
Let's call :
$$\overline{x_1x_2} = a$$
$$\overline{x_4x_5x_6} = b$$
$$\overline{x_8x_9} = c$$
So
$$a \cdot x_3 = b = x_7 \cdot c $$


*

*What we know:

*

*$x_3$ or $x_7$ cant't be $1$ 

*$a$ or $x_3$ can't be both primes. Also $c$ and $x_7$.

*$b$ cant't be prime

*$ 123\le c\le 776 = 98 *7 $

*Last digit of $a$ or $c$ can't be $1$ because $\color{red}{x_3} \cdot \overline{x_1 1} = \overline{x_4x_5 \color{red}{x_3}}$

*The last digit of a,b or c can't be $5$
This narows c to 500 numbers. If i have more ideas i will post here.


