Piccadilly is primarily a mathematical setter these days, and his last Listener was three years ago with the superb Properties of Numbers — II. I’m not sure how setters go about clueing this sort of puzzle we had this week. Do they think of arbitrary words or phrases that just appeal to them or is there something more scientific? Whichever, R^A – T + (E+N-T+E-R+S)^A(BIN – N) was probably my favourite!

The straightforward way into this puzzle was through the last clue 29dn T^A which could have few possibilities given that it was one of the three clues that we were told was palindromic in the preamble. T! in 14ac excluded 11², so 7³ it was with T = 7, A = 3. 21ac TA + R^R + IE + R + S pretty much gave R = 5.

From there, the solution came fairly quickly with no more than a calculator required, although I’m sure some solvers would have created a spreadsheet with macros to solve it — just for fun. I’m not one of those. Once the grid was filled, converting the numbers to letters gave I hate numerical crosswords because making a mistake means starting all over again. Luckily, I wasn’t required to do any backtracking here, so breathed a sigh of relief.

However, I felt that this surely didn’t describe Piccadilly’s attitude to his artwork! I wondered whether there was an alternative, such as I love numerical crosswords despite making a mistake meaning a start all over again! Sadly no.

Very enjoyable, thanks Piccadilly.
 

There is one thing I regularly say about numerical puzzles: I HATE NUMERICAL CROSSWORDS BECAUSE MAKING A MISTAKE MEANS STARTING ALL OVER AGAIN. We (or I should say ‘he’ as this three-monthly departure from the comfort of the verbal ones is his province) invariably make that mistake and the air becomes somewhat ‘blue’ at the third (or tenth) new start. This time, after a couple of hours fumbling and muttering, he announced “Impossible! I can’t do this with just pencil and paper.” (which is his approach to the things).

343 went in fairly early on at 29d, 7 cubed and a palindrome, so T = 7. That was about the limit of my comprehension but I know Tim (Encota) will be providing a detailed explanation of his progress so should leave that to him.

My other concern with these three-monthly departures is whether we can admit numerical setters to the Elite Listener Oenophile Outfit. If there are no doubles or triples, it becomes rather dicey. But there was hope for Piccadilly. All his clues made real words (quite a feat in itself). Initially I was feeling some ANGST! (14ac) and GOING BANANAS (25ac) and together with ROUT, CARNIVALS, RAT ENTERS A BINN and USE BEAN TINS (!) was wondering what was going on, but Piccadilly redeemed himself. There was MALT at 35ac. What can I say? 39d TA Piccadilly.

Postscript: Piccadilly, if you hate them as much as I do, what are you doing setting the things? If the editors received none at all, they might treat us to misprints, jumbles, Playfair codes and all those other joys.

Another postscript: A solver has sent me his solution route which turns out to be a lot more straightforward than we thought.

14a => T! < 10000 => T < 8

29a is a 3-digit palindrome => either 11^2 or 7^3 so T=7, A = 3, 29d = 3-4-3

30a => S < A => S = 1 or 2.  But S=2 => 30a = W – O + K cannot be 3-digit => S = 1

22d => R>A  => AR is at least 3*4; 17a => C^AR < 100000 => C < 12^th root of 100000 ~ 2.6 => C=2

4^4 = 256, 6^6 = 46656. So the only way 21a can be 4-digit is R=5

31a => E! <1000 => E<7; 17d = 2^E(5 – 1 + M + I +2) – 1.  If E were 4, that would be at most 2^4(6 + 18 + 19) < 1000, so E > 4 => E = 6

I > G.  WAN = 3WN is at most 3*18*19 < 3*20*20 = 1200.  If I were 9 or bigger, A^I would be at least 19683, and 4d would be 5 digits; thus 19683 I=8, G=4, 21a=3-19-8

8d is a palindrome divisible by G=4 with one cell being 2-digit => 8d = 2112 or 6116.   8d = 4L(OB + U – L – 6) > 4*9(9*9 + 9) = 3240 => 8d = 6116 (either 6-11-6 or 6-1-16)

So L divides 6116/4 = 1539 => L = 11

34a = 8N^2 + 5 < 1000 =>8N^2 < 995 => N^2 < 122 =< N < 12.  L=11. So N=9 or 10.  If N=10, 34a = 805  and there are no cells with 0.  So N=9, 34a = 6-5-3, 14a=5-15-1

32a = COOL – S = 22*O^2-1 is 4-digit beginning with 4 => 4000 < 22*O^2 < 5000 => 181 < O^2 < 227 => O = 14 or 15.  If O=15, 32a=4949 which cannot be entered as 3 cells.  O=14, 32a=4-3-11, 26d=18-5-6, 7a=19-6-5

7d = 53*M^2 is 5-digit starting with 19 => 358<M^2<377 => M=19, 7d=19-1-3-3, 4a=6-9-16, 35a=4-3-8-9, 9d=5-13-1, 11d=2-5-19, 27d=19-8-3

22d = 2*B^3-7 = 19_3 from the crossing entries => B^3 <= 1000 => B=10, 22d=19-9-3, 10a=8-2-3, 18a=18-8-9-4, 25a=7-3-18, 1d=8-8-14-6, 3d=7-11-1-18

8d=6116 from above means that U=16, 13a=1-1-13 (which resolves 8d=6-1-16), 16a=6-19-3, 24d=1-1-11, 5d=9-5-6-9, 6d=16-14-2

From crossings, 15a=2-3-16 => V=17, 17d=3-3-3-17, 19d=8-6-7-5, 23d=8-1-11-9

28a = 9*(3K + 17) = _39 from crossings so must 639 =>K=18, 28a=6-3-9

30a =W^2 +4 = 1_3 from crossings => W^2 = 1_9 => W=13, 30a = 1-7-3, 20d=4-6-9-4. 4d=6-2-10

That leaves D=15 with the clues using D confirmed by the full grid.