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A few years ago, my child brought a puzzle home from their 6th-grade math teacher. The solution was to use all the numbers $1−9$ only once and make the equation true. Using a spreadsheet, I was able to calculate the 2 answers: $631892754$ and $631472958$

My problem is I found the spreadsheet and the answers but I don’t remember the original problem. I have tried to recreate it using my spreadsheet formulas, but just not getting anywhere.

I think it was in a format similar to _ _ * _ _ * _ = _ _ ∗_ _ (each underscore a different number $1-9$) Might even have some parenthesizes. I don’t think it had any division, but I could be wrong. I am also fairly certain it was a single equation. I remember showing it to a co-worker the next day and I was able to write the question down from memory. It wasn’t overly complicated since I was able to figure out quickly what was required and it didn’t require any advanced math.

These are my formulas from the spreadsheet for the first solution $631892754$.

Step 1: Using all combinations of digits 1, 2, and 5.

Digit $1$ multiplied by digit $5$, divide by digit $2$. Result is digits $3$ and $4$.

Answer must be a whole number greater than 10, cannot end in 0.
Using the first answer $(6∗9)/3=18$

Step 2: Using the digits and the answers from Step 1.

Digit $5$ multiplied by digit $1$ must equal digits $3$ and $4$ multiplied by digit $2$
Using the first answer $9∗6=18∗3$

Step 3: I have 5 numbers, determine which numbers have not been used for each potential answer. Tried all combinations of the remaining 4 digits until I got the answer.

Digit $6$ and $7$ multiplied by digit $1$ must equal digits $8$ and $9$ multiplied by digit $2$
$27∗6=54∗3$

I spent a few hours searching the Internet for this puzzle, but no luck. I am hoping the above information is enough to go on.

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  • $\begingroup$ computers allowed? or is there a logival solution? $\endgroup$
    – Jasen
    Oct 23, 2021 at 3:26
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    $\begingroup$ Step 2 doesn't add any constraints. $\endgroup$
    – Muqo
    Oct 23, 2021 at 8:57
  • $\begingroup$ This is truly a question not a riddle or similar. I know the answer and I am looking for the original question. I have no limitations on what tools you can use, including a computer. $\endgroup$ Oct 23, 2021 at 20:40

2 Answers 2

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I'm going against some of the things you suggested here but one possibility I've found that could work is the following (where each space must be filled by a unique digit between 1 and 9)

_ ÷ _ = _ _ ÷ _ = _ _ ÷ _ _

Framing it in this way makes it look like a single equation. I checked using a computer and this has exactly two solutions

6 ÷ 3 = 14 ÷ 7 = 58 ÷ 29
6 ÷ 3 = 18 ÷ 9 = 54 ÷ 27

which look like yours but with the end pairs switched.

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  • $\begingroup$ @Heximo: thank you for the equation. It very well could be an equation with 2 equals. That would explain the math I was doing. I am thinking the last set of numbers are just reversed because the way I rewrote the equation to do the calculations. Thank you! $\endgroup$ Oct 23, 2021 at 21:24
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Building on hexomino's idea and doing a different kind of cheat from the two equal signs thing:

$\sqrt{(6 ÷ 3)(18 ÷ 9)}\cdot27 = 54$
$\sqrt{(6 ÷ 3)(14 ÷ 7)}\cdot29 = 58$

However, a quick python program reveals that there are more solutions than just those two:

1 3 9 6 8 2 7 5 4
1 6 2 4 9 5 7 3 8
1 7 2 8 9 5 4 3 6
1 8 9 6 3 2 7 5 4
1 9 2 4 6 5 7 3 8
1 9 2 8 7 5 4 3 6
2 4 5 6 7 1 9 3 8
2 7 5 6 4 1 9 3 8
4 2 3 6 8 1 9 5 7
4 8 3 6 2 1 9 5 7
6 3 1 4 7 2 9 5 8
6 3 1 8 9 2 7 5 4
6 7 1 4 3 2 9 5 8
6 8 2 7 9 3 4 5 1
6 9 1 8 3 2 7 5 4
6 9 2 7 8 3 4 5 1
7 3 8 4 9 1 2 5 6
9 3 5 4 8 1 6 7 2
9 8 5 4 3 1 6 7 2

However, these don't all solve the equations in the OP's 1 (=2) and 3. Hexomino solved both equations for $d_1/d_2$ which matches the constraints and the solution, so seems close, although the swapping of the last two pairs seems odd (equation 3 is $\frac{d_1}{d_2}=\frac{d_8d_9}{d_6d_7}$).

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