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Can you place all numbers from 1 to 9 into cells, such that the following 5 equations hold? Good luck!

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Answer

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Step-by-step walkthrough

Consider the third row. One of the multiplicands cannot be 1, or else the other two numbers would be the same. Likewise both multiplicands cannot be 3 or greater (3x3=9 is a repeat; anything else is greater than 9), so one of the multiplicands is 2. This means that the third row must be (2,3,6) or (2,4,8) where we can swap the first two digits.

Next, consider the first two rows. Notice that in each one, the sum of the digits in the row must be even. This is because for the first one where A-B=C, this means A=B+C, so therefore A+B+C=2(B+C). Likewise the second one's sum is twice the 3rd digit.
Since the total sum of the digits is 1+2+...+9=45, the third row's sum must be odd. So therefore the third row must be (2,3,6) or (3,2,6).

The remaining digits are (1,4,5,7,8,9). One of the rows doesn't contain the 1, and since the row is either a sum or a difference (which can be rearranged to a sum), the addends are at least 4 and 5, which makes the sum at least 9, which is a tight bound. So one row contains the digits (1,7,8) and the other contains (4,5,9).

I don't have an extremely clean way to proceed from here, but fortunately there are not many cases to check now. Suppose (1,7,8) is the top row. Then the 8 is in the top left corner, the bottom left corner must be a 3 (if it were a 2, that makes left middle = 6, but 6 is already used), so the center column is now forced to be X-4=2, and this is bad.
The other case is if (4,5,9) is on top. 9 is in the top left, thus 2 is in bottom left (3 bottom left means 6 middle left) and therefore the center column is X-1=3. So X=4, and the rest of the grid can be uniquely filled.

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    $\begingroup$ Just beat me to it. Another way to explain the last step is to check the possibilities above 2, which can be only (7,5) or (9,7). If we choose (7,5), it must be on the second column because both are addends, but then we get a contradiction on the first column. $\endgroup$
    – Bubbler
    Oct 7 '20 at 1:15
  • $\begingroup$ Well done you got it! You can now try the 4x4 version of the puzzle, which I posted. $\endgroup$ Oct 7 '20 at 2:09

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