I recently ran across this puzzle: Place the numbers 1, 2, 3, 4, 5, 6, 7, 8 in this gridenter image description here

So that each row and column of 3 digits sum to the same number. I was able to find a number of solutions. The interesting thing was that some of the solutions were related to each other. Besides the rotational symmetry, there is the property that given one solution, if you replace each number $a$ by $9-a$, you get another solution. There seems to be a more subtle relationship between some solutions but I haven't yet nailed it down. I don't want to spoil, so I have to be vague. It looks like if a column is $a, b, c$ in a solution, then there is another solution with $a, c, b$, and the other 3 rows permuting similarly.

My question here is really to ask for references to this puzzle. I know I just ran across it in the last month or so, but I can't find it on Google or StackExchange or anywhere. I'm curious as to whether one solution leads to all the others by rotation, by the $9-a$ rule or by some permutation trick.

If anyone cares, I can share some of my solutions and the reasoning that led to them.

EDIT: As I said, I have solutions, so I'm not asking for solutions, but relationships between the solutions. The $9-a$ trick is an example. I think there is some permutation trick that will turn some solutions into others, as I mention in the comment below.

  • $\begingroup$ @hexomino Yes, thanks. I must have seen the puzzle here. In your answer there, you list 12 solutions. By the $9-a$ trick, these could be paired and we could say there were only 6 different solutions. Your first and last solutions are related this way. Notice that in your fourth solution, each row and column is a permutation of a row and column in your first solution. I'd like to work out a rule that converts your 1st solution to your 4th, and also groups others. Maybe there's really only one solution? $\endgroup$
    – B. Goddard
    Aug 28, 2020 at 13:30
  • $\begingroup$ I think you must be talking about Alexander's solutions, correct? Yes, it does look like we can get down to 6 by using the $9-a$ pairing. Let's say I then choose my six to be the four with 1 on the outside and the two with 3 on the outside. It looks like these can be further formed into 3 pairs by saying that pairs essentially have the same rows and columns but permuted. However, I would still count these as distinct so I wouldn't go further than this (down to 6, but not 3). It depends on your perspective though. $\endgroup$
    – hexomino
    Aug 28, 2020 at 13:43


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