3
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I made a puzzle. I don't think it should be too hard.

Puzzle involving placing digits in squares

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  • 1
    $\begingroup$ Even without the last rule, this still has a unique solution. $\endgroup$ – Kruga Feb 5 '18 at 9:05
  • $\begingroup$ Fixed! For future reference, the last rule was that two edge adjacent squares must have product less than 50. $\endgroup$ – Martin Feb 5 '18 at 12:38
  • $\begingroup$ What is the definition of a digit with a loop? $\endgroup$ – Ben Mordecai Mar 16 '18 at 3:25
3
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2 9 5
6 1 8
7 4 3

Here's my reasoning:

1 is not prime and has no loop so must go in the centre. The corners are the primes 2357, the edge cells are the rest, 4689. The rows must add to 16, 15, and 14 because the total of all three rows is 1+2+..+8+9=45. The only possibilities for a 15 sum in the middle row is 816 or 618. This leaves as the only possibility for the 16 sum in the top row 295. The bottom row is then 743 or 347. A column of 567, 278, or 578 would sum to more than 16, leaving only one possibility.

As Kruga pointed out, the last rule (about adjacent squares multiplying to less than 50) is unnecessary. I have edited my explanation to avoid needing that rule.

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