<s>[INST] $\square\square\times\square =\square\square\square =\square\times\square\square\,\,\,$ 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.
 [/INST] 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.
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