Comments on: A shy woodland creature

By: Delmania

Delmania — Thu, 27 May 2010 18:06:51 +0000

If I am reading this right, you are removing all the numbers in the same row and column as the one you circle. In that case, I think that leaves on diagonals, in which case, the sum of the diagonals are all 57.

By: bach

bach — Tue, 25 May 2010 21:27:19 +0000

I think it is easiest to observe each row can be obtained by adding a constant to the entries of the first row: for example, the second row is the first row minus 7. So if we label the entries in the first row as x1, …, x5, and the constants as c1, …, c4, any combination of 5 values chosen from distinct rows and columns has to sum to (x1 + … + x5) + (c1 + … + c4).

By: leonardo

leonardo — Tue, 25 May 2010 11:18:02 +0000

As a good mother, Martin has created numerous mathematicians that have discovered many things. It’s a good way to meta-invent mathematics.

When you have time you can show us more article scans of very old computer magazines and the like.

By: Nemo

Nemo — Tue, 25 May 2010 05:04:05 +0000

Yeah, very pretty.

I started with a Lemma:

Suppose the square “works”. Pick any four cells from the grid that form a rectangle. Diagram:

a … b
…
c … d

Then a+d = b+c. (Because if you circled a and d, you could have circled b and c instead and crossed off the same rows and columns.) So this condition is necessary for the square to work.

But it is also sufficient. Given a square that satisfies the condition and any five circled numbers in different rows/columns, choose b on column N and c on row N, labelled as in the diagram above. (Such b and c exist because every row and column must be used.) Then replace the circled pair (b,c) with the circled pair (a,d). Observe that d is at row N and column N, and is therefore on the diagonal. Proceed with such replacements until all five circled numbers are on the diagonal. Since each replacement did not change the sum, the sum of original five circled numbers equals the sum of the diagonal.

…

From there it is a quick step to svat’s formulation concerning the differences between adjacent rows/columns.

By: svat

svat — Tue, 25 May 2010 03:48:42 +0000

Beautiful! We will miss Martin Gardner’s genius…
It had me amazed for quite a while… there are 120 possible sums that must all be equal, with only 25 numbers to choose?

But after working it out, though, it seems so natural. It can be no other way; the differences have to be equal. And the same idea works for square matrices of any size!

I also started wondering whether, since we have essentially only 9 “free” choices, we could make all the numbers distinct without making them too large… then I was embarrassed to realise that simply writing the numbers 1 to 25 in order is such a matrix. :-)