# How many distinct solutions are there?

Suppose you put the numbers $1,2,\cdots ,10$ in each of the boxes below

such that every connected row and column sum to the same number. How many distinct solutions are there? (By distinct we mean disregarding cyclic permutations, reflections, etc.)

I tried this problem and found only one solution, namely $4,5, 1,10,8,2,3,6,9,7$ when read clockwise starting on the top leftmost square. I tried to prove it's unique but failed (miserably).

From experience, I know algebra isn't the best way to approach these types of problems, due to the number of variables and the symmetry of the equations. When trying to find $3\times3$ magic squares, for example, it'd be best to brute force all the possibilities rather than trying to solve a $9$-variable system.

So does anyone know how to solve this (preferably without guess-and-check)? Any help is welcomed and appreciated.

• "How many unique solutions are there?" Just one. If there's more than one, then it's not unique. – bof Oct 31 '14 at 1:09
• I think the word you're looking for is "different" (or "distinct" or "distinguishable"). – bof Oct 31 '14 at 1:10
• Here is another solution: $3,1,10,5,8,6,4,2,7,9$. – David Oct 31 '14 at 1:11
• @bof I agree with you: unfortunately nowadays the word "unique" is often used with this incorrect meaning, particularly in the expression "this site has had $n$ unique visitors since $d.m.y$." – David Oct 31 '14 at 1:14
• A couple of more or less obvious observations: $10$ and $9$ cannot be in the center of either row, since then the row sums must be at least $22$ and there is then no way to get both columns to sum to $22$. Second, the sum of the numbers in the middle of the two columns is odd; by the first point, it must be at least $13$, so it is $13$, $15$, $17$, or $19$. – rogerl Oct 31 '14 at 1:30

This is not a complete answer but just a few observations which are too lengthy for a comment.

Let $t$ be the total of the numbers on any one side of the rectangle, let $c$ be the sum of the four numbers in the corners, let $s$ be the sum of the two numbers in the middle of the short sides (columns) and let $l$ be the sum of the four numbers in the middle of the long sides (rows). Then

• $c+s+l=55$;
• $3c+2l+s=6t=3c+2s+l$, so $s=l$;
• $4t=55+c$, so $c\equiv1\pmod4$;
• $2t+l=55$, so $l$ is odd.

Taking into account the possible ranges of the variables, the options we have at this stage are $$\matrix{ c={}&13&17&21&25&29&33\cr t={}&17&18&19&20&21&22\cr s=l={}&21&19&17&15&13&11\cr}$$

There are ten solutions \begin{align*} &1,4,5,9,7,3,2,6,8,10 \\ &1,4,7,6,10,2,3,5,8,9 \\ &1,5,6,7,9,3,2,4,10,8 \\ &1,5,7,6,9,4,2,3,10,8 \\ &2,3,6,9,7,4,1,5,10,8 \\ &2,3,8,5,9,4,1,7,6,10 \\ &3,1,10,5,8,6,2,4,7,9 \\ &3,2,5,10,6,4,1,7,8,9 \\ &6,1,5,10,4,8,2,3,9,7 \\ &6,3,5,8,4,10,1,2,9,7 \end{align*} which turn into $40$ distinct solutions when the numbers in the middles of the upper and/or lower rows are reversed. These are all the solutions, regarding rotations and reflections as the same.

• I'm not really interested in the solutions themselves, but how you get them. Like I said, this would help me in many other problems similar to this one. – Edward Jiang Oct 31 '14 at 2:35