Place numbers 1 through 9 in boxes (☐☐×☐=☐☐+☐☐=☐☐)

Question

So recently I was scrolling through Youtube when I came across this video from MindYourDecisions that was about solving a legendary math puzzle.

The puzzle:

Place the numbers 1 through 9 in the following boxes such that

 ☐☐
 x☐
 ---
 ☐☐
+☐☐
 ---
 ☐☐

Each number can only be used once.

Here is the thumbnail of the Youtube video for anyone who wants to see what it looks like:

Now, I recognized this puzzle as one that I have been struggling to solve on and off for over a decade, and that is bad because I still have not managed to solve this.

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What I can really easily deduce

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1. If we have 9 in the topmost box in the left column, then only if we multiply by 1 will we be able to satisfy the condition that we have a 2-digit number multiplied by a 1-digit number that is then equal to a 2-digit number. But, this leads to a contradiction because we have to use every number once.
2. We can use the same logic to show that 9 cannot go in the 2nd and 3rd topmost boxes, because we require that we have 2 2-digit numbers being added to each other that is equal to another 2-digit number. However, this cannot be satisfied if we have 9A+BC (with AB≠A*B, but rather representing concatenation).
3. We can also use the same logic as 1) to show that 5, 6, 7, and 8, also cannot go in the topmost box in the left column.
4. We can notice that since the multiplier cannot be 1, then it follows that there cannot be a 1 in the second topmost box or the bottom box in the left column.
5. We also cannot have the numbers 8 or 9 as the multiplier, as the minimum value given from any multiplication would result in 96. We can also probably eliminate 6 and 7 from being the multiplier, although I'm not too sure if that would be correct to do.

So this is what I have done so far (a blue number means that I am 100% sure it does not go there, a red number means that I am pretty (but not 100%) sure that it could not go there):

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However, I am actually unsure about how I would deduce logically (by hand) where the numbers go, so my question is:

What should I do to actually solve this?

_{Edit: Sorry, forgot to mention that I am looking for a logical way to hand-solve this.}

Answer 1

I would focus on the single digit number

X * (6+) Total not possible 13*6+24 >98)
X * (5 or 1) Last digit in product not available
X * 4:
X not 19+ Total not possible $19 * 4+23>98$
X not 11,14,15 or 16 Last digit in product not available
$12*4$ First digit in product not available
$18*4$ ends on 2 Total not possible $18*4+35>98$
$13*4 =52$ Total not possible $13*4+67>98$
Only $17 * 4$ left to check: Solution! $17*4 =68+25=93$

X * 3:
X * 2:
Still a lot of possibilities to check for uniqueness. I don't see an obvious overall smart way to do that.
though e.g.:
$4?*2 =?? +??=??$ ->
$4?*2 =8? +1?=9?$ ->
$43*2 =86 +1?=9?$ (6 is only remaining even number) -> no solution with $4?*2$