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The numbers {1 to 11}, excluding 8 and 10, are arranged in a 3 x 3 matrix.

The bottom row sums to 22. The rightmost column sums to 24. The back diagonal (/) values sum to 14.

Each triplet of values within a row, column or diagonal are pairwise coprime. That means that any pair of two values of out the three have a greatest common divisor of 1.

Find the numbers!

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We have 3 even numbers to insert. Since the numbers in rows, columns and diagonals are coprime, it means that the even numbers must have different row, different column and different diagonal. Also, we have to manage the 3 even sums; a even sum can be made with E+E+E or E+O+O (where E=even and O=odd). The E+E+E case is not possible, as we already stated that the even numbers have different rows and columns. Now we know that each sum (14,22 and 24) contain only one even number. Now we have few cases:

  • Bottom-Right cell is even. Though, this excludes all the / diagonal cells from being even, which is wrong as 14 cannot be obtained as sum of 3 odd numbers. Contraddiction!
  • Middle-Right cell is even. This excludes the Top-Left and the Middle cells from being even too, consequently Bottom-Left must be even! The third even cell is Middle-Top.
  • Top-Right is even. Using the same logic as before, we deduce that Bottom-Middle must be even, as well as Left-Middle.

In the right column, the even number must be 6, else we wouldn't find partners for the sum. 24-6=18. The only way of making 18 is 11+7, so the right column must be a permutation of {6,7,11}, 6 not being in the bottom-right corner. 7 cannot be the bottom-right, else we wouldn't find partners for 22 (remember that one must be even).
So, 11 is forced to be the bottom-right number! The bottom row can either be a permutation of {11,2,9} or {11,4,7}; though, we already used the 7, so the only possible bottom is permutation of {2,9,11}. If 9 was the bottom-left, we would have serious problems with the diagonal sum 14, so the number 2 is in the bottom-left corner, forcing 6 to be middle-right and 4 to be top-middle. Current situation is
? 4 7
? 5 6
2 9 11

3 cannot be in the 2nd row because 6 is present, so the final answer is:
3 4 7
1 5 6
2 9 11

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    $\begingroup$ That was a huge logic loadout, but somehow I did it! $\endgroup$
    – leoll2
    Commented Apr 13, 2015 at 20:01
  • $\begingroup$ Well done for a good explanation of your logic @leoll2 $\endgroup$
    – Paul Richards
    Commented Apr 13, 2015 at 20:08

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