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

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.

### What I can really easily deduce

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):

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.

• Without no-computers it's trivial to brute force. Do you just want the numerical solution, or are you specifically after a logical way to hand-solve this?
– fljx
Feb 28 at 15:57
• FYI, the video's solution method is not elegant, as it was essentially testing a huge number of cases, only one of which leads to a solution, to the point where one is just emulating a computer's brute force search. Feb 28 at 16:53
• You also can't have 5s in the top three spots of the right column. Feb 28 at 17:24
• FWIW case bashing to death is a kind of logic. Some "higher-level logic" can help you reduce the amount of work, but I believe you'll need to weed through at least some tens of cases in the end. Feb 29 at 0:56
• @loopywalt You're right. For some reason I misread the entire thing as a long multiplication summing two products to a final answer (but that would require a two-digit multiplier, and the answer would never be only two digits). Not sure what I was thinking :-/
– fljx
Feb 29 at 11:35

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 $$4?*2$$