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Solution 27 34 41 48 19 31 43 55 11 28 45 62 3 25 47 69 Reasoning Filling in the first column as an arithmetic progression and then the rows as arithmetic progressions we get...
#1: Initial revision
by
Peter Taylor
·
2026-08-03T08:08:46Z (about 2 months ago)
<details><summary>Solution</summary> ```txt 27 34 41 48 19 31 43 55 11 28 45 62 3 25 47 69 ``` </details> <details><summary>Reasoning</summary> Filling in the first column as an arithmetic progression and then the rows as arithmetic progressions we get ```txt 27 27+b 27+2b 27+3b 27+a 27+a+c 27+a+2c 27+a+3c 27+2a 27+2a+d 27+2a+2d 27+2a+3d 27+3a 27+3a+e 27+3a+2e 27+3a+3e ``` We have five unknowns. Taking any of the other columns as an arithmetic progression we get $c-2d+e = 0$ and $b-2c+d = 0$; together with the three other given values we have five linear equations in five unknowns and we can apply Gaussian elimination. The stated inspiration is a hint that this was the intended solution process. It could also have been done with twelve variables, one for each empty square, but it would have been less straightforward to identify the redundant equations, and they would also have been more fiddly to manipulate. </details>