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This puzzle is part of the Monthly Topic Challenge #14: Think inside the (very small) box!.

Place the digits 1 to 9 in the box below in such a way that in each row, each column and each diagonal, one of the numbers is the sum of the two others.

a 3x3 box with digits

Notes:

  • Combining digits is not allowed. Put one digit in each cell.
  • Rotations and symmetries don't make a difference.
  • The digits in the image are there for illustration purpose. Ignore them.

While this task might look impossible at first, think harder.

Hint:

Taken at face value the task is indeed impossible. You need to rely on some lateral thinking. Almost cheating.

My apologies for that.

(this is my own creation)

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  • $\begingroup$ Are the numbers (1,2,3,5,8,9) aside the grid useful for the puzzle or pure decoration ? $\endgroup$
    – Evargalo
    Sep 7 at 12:38
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    $\begingroup$ No, these are just for the illustration, floating around, ready to be placed. I updated the question. $\endgroup$
    – Florian F
    Sep 7 at 12:39

3 Answers 3

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I placed one of each digit in the squares:

165/628/743

It looks like all of the conditions are met.

Oops, the 9 got rotated a little bit.... : )

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  • $\begingroup$ Damn, that was fast. That was the twist I had in mind. Bravo. $\endgroup$
    – Florian F
    Sep 7 at 14:42
  • $\begingroup$ @FlorianF: Haha, rot13(gur gjvfg vf... n yvgreny gjvfg)! $\endgroup$
    – psmears
    Sep 8 at 14:57
  • $\begingroup$ I actually interpreted the "no rotations" comment in the question to preclude this solution, but now I see it was talking about grid rotations. $\endgroup$
    – GentlePurpleRain
    Sep 20 at 20:26
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All digits once, no rotations:

enter image description here

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  • $\begingroup$ Nice solution! I didn't think of adding math symbols. So that isn't the expected solution. But besides that, it seems to fit the bill. $\endgroup$
    – Florian F
    Sep 7 at 14:42
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    $\begingroup$ Comment by Kilgorezer (who doesn't have the rep to comment): Amoz's and Retudin's solutions are actually the same, just written differently. I didn't notice that. $\endgroup$
    – Florian F
    Sep 9 at 13:49
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There's also another solution:

Just put an 8 at the center and rotate it 90 degrees. That leaves 3 odd and 5 even numbers:

abc
d_e
ghi

On each of the two outer parallel rows (and columns), there must be an even number of odd digits, leaving an odd number of even ones. Only with one even digit on each outer row (or column for that matter) is it possible. Also, one of these even digits must be on a corner and the others must be an L-shape away from it ($a$, $e$ and $h$ for example). Let $a=2$:

253
9_4
761

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  • $\begingroup$ As long as the four lines not involving the middle square also satisfy the criteria, of course... $\endgroup$
    – Stiv
    Sep 13 at 11:58
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    $\begingroup$ Now it's complete. $\endgroup$
    – Nautilus
    Sep 14 at 9:15

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