## Listener No. 4399: *Square Time Sums* by Oyler

Posted by Dave Hennings on 10 Jun 2016

It had been over two years since Oyler’s last puzzle with its University of St Andrews theme. Before that was *2x2x2* which required us to cut out four cubes and arrange them into a larger cube in order to complete the dice net required for entry. That endgame was certainly tricky, and I hoped this week’s endgame was a bit easier. The fact that it was Oyler’s 13th Listener had me concerned!

Here we had to identify four triples, each consisting of 2-, 3- and 4-digit squares using all the digits 1-9. My first step was to pencil in the last cells of all the squares with 1, 4, 5, 6, 9 and 1, 3, 7, 9 for any entry given as a prime.

After that, 1dn *7 x X* (3) seemed a good place to start since X had to be 16-81 and 7 x X provided the first digits of J, K and L so its digits had to be different. Only 175 and 567 filled the bill, but there is no 2-digit square 5?, so 175 got slotted in. 1ac was therefore 16. 2dn was prime, but there is no prime 62? so 2dn was 683 and 15ac was 5329. (J, K, L) was thus (16, 784, 5329).

Unfortunately, the rest of the solve went a bit more slowly and, in some cases, required the teasing out of single digits or the gradual rubbing out of pencilled options. I will leave the detailed solution to other bloggers or to listenercrossword.com, but in summary:

(A, B, C) was (36, 729, 5184)

(J, K, L) was (16, 784, 5329)

(P, Q, R) was (81, 576, 3249)

(X, Y, Z) was (25, 841, 7396)

And so to the endgame which required the highlighting of two rows and two columns which were related to the theme. 13 be damned! The 6th and 9th rows and columns each contained all the digits 1-9 and were also squares.

254871369 = 15963

847159236 = 29106

184231675 = 24807

412739856 = 20316

It wasn’t too difficult a puzzle after all, but enjoyable nonetheless. Thanks, Oyler.

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