Listen With Others

Are you sitting comfortably? Then we’ll begin

Posts Tagged ‘Piccadilly’

L4725: ‘Carte Blanche’ by Piccadilly

Posted by Encota on 9 Sep 2022

Great fun! I always love it when a Numerical appears by Piccadilly. The puzzles entitled ‘The Properties Of Numbers’ in earlier Listeners were an absolute delight – perhaps my favourite ever numericals – so this puzzle immediately got my interest.

As ever it needed careful concentration throughout to avoid any simple mistakes creeping in. But there was enough cross-checking to ensure (I hope!) an error-free gridfill. Here is my effort.

I hope you enjoyed it as much as I did!


Tim / Encota


Posted in Solving Blogs | Tagged: , , | Leave a Comment »

Carte Blanche by Piccadilly

Posted by shirleycurran on 9 Sep 2022

I really loathe the numerical crosswords and invariably plan not to write a solver’s blog this time but then we bungle our way through them – this time it took us forever to sort out those clue numbers and begin, and we managed to bungle right royally with V, J and Q creating clashes at the end but a friend who actually seems to enjoy them has sent me his solving notes with permission to use them, so here goes!

“There are 43 clues in all, so no clue number is more than 43.

Down SO & HO => S < H

Across HE & YE => H < Y

If S>1, then (SH)^Y is at least 6^4 which is much too big.  S=1

If H>2, then (SH)^Y is at least 3^4 which is too big. H=2

O-R is at least 8 since it’s the 8th down clue.  If R is 4 or more, then O is at least 12 => OR is ate least 48 which is too big.  R=3

Down HUH  = 4U < YU => Y > 4.  If Y >5, (SH)^Y is at least 64.  Y=5

R + YE = 3 + 5E.  If E > 8, that would be at least 48, so E < 9.  O-R is at least 8, so O is at least 11; HO is at least 22.  So ER = 3E > 22.  E = 8  
HO < 24 => O < 12.  O=11.   
O-R = 8, so the first eight downs are numbered 1 through.  This fixes U=7, D=13, T=22

OR = 33, YU=35 so P+O=34  P=23

HI < HO so I < 11.  S+I-N = 4, so I = N+3.  The only unused values less than 11 are 4, 6, 9, 10.  N=6, I=9

A+S>ER=24 So A is 24, 25 or 26.  A-M = 6.  So M = 18, 19 or 20. A+N < (SH)^Y=32, So A is not 26.  If A = 24, the across numbers go RY+E=23, AS=24, RI=27 and the downs go ER=24, A+S=25, HUH=28.  Then there would be no clue 26.  Thus A=25, M=19

Across numbers go A-N=19, S+I+C, UR=21.  So S+I+C=20.  C=10

C+O-B = 5.  B=16

Downs numbers then go HUH=28, M+O=30, so there must be 29 across.  RI=27, T+O=33 so 29 is either G+I or G+O.  If G+O = 29, then G=18 and G+9 = 27 But that’s RI.  So G+I=29.  G=20

YE=40 and R+YE=43 so RI+SK=41 and F+O+B=42.  K=14F=15

Across numbers go E-R=5, I+S=10, so there must be 9 down.  C+O-L = 9.  L=12

H+A-W < 5 so W > 22.  W = 24 or 26 as 23 and 25 are used.

If W = 24, then H+A-W = 3.  Then we would have 3acr=SO^Y=161051 and 3dn = S/AY = 0.008 which do not agree.  W=26.  (You knew the actual entries would need to be used at some point😊)

All of the numbers are now used except 4, 17, 18, 21, 24 for J, Q, V, X, Z in some order.

5acr = (Z+E)^N = (Z+8)^6  If Z were 17 or more, this would need at least 9 digits.  However, symmetry means 42acr = (E-N)V^Y= 2*V^5 must have the same length.   2*24^5 has only 8 digits.  So Z = 4

We now have all letters except J, Q, V and X and all entries except 12acr, 19acr, 29acr, 39acr, 42acr, 22dn, 26dn, 30dn, and 31d.  It’s pretty easy then to determine that the grid fill must look like:

12acr, 19acr, 29acr and 22dn can only work if Q=18, X=17, V=21 & J=24

Many thanks to my clever and kind friend! Our grid looked like that, and I would have left it at that except that, of course, I had to check how Piccadilly fares in the Listener Oenophiles outfit – disppointment, his clues were not swimming in alcohol. But then, to my amazement, I saw that the DIZZY HARES had danced into the clues (probably chasing the DIRTY ROTTER – Piccadilly must have. had tremendous fun creating all those real-word clues).

Dizzy Hares

A final comment. The other Numpty has actually admitted that he admired and enjoyed this one so much that if (by some miracle) we managed an all correct this year, it would certainly be in his five top selected crosswords.

Posted in Solving Blogs | Tagged: | 1 Comment »

Listener No 4621: True by Piccadilly

Posted by Dave Hennings on 11 Sep 2020

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, RA – 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 TA 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 + RR + 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.

Posted in Solving Blogs | Tagged: , | Leave a Comment »

True by Piccadilly

Posted by shirleycurran on 11 Sep 2020

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 < 12th 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.

Posted in Solving Blogs | Tagged: , | Leave a Comment »

The Properties of Numbers II by Piccadilly

Posted by shirleycurran on 8 Sep 2017

I downloaded a rather large symmetrical grid with the usual anxiety I feel when the three-monthly numerical crossword appears. My trepidation only increased when I read that more than half the clues consisted of digit sums or simple additions of other solutions and not of mathematical information that could give us a tiny claw hold as a way in. “20 = 10 + 10”. Well, I know that and all the other simple addition and multiplication facts like 6 = 3 x 2 but it wasn’t so simple was it? Though it was indeed admirable how Piccadilly made all those mathematical basics correspond with the reality of his far more complex mathematics. He didn’t convince me, though, of his right to retain his admission ticket to the Listener setters’ winos outfit as I couldn’t find any alcohol in his grid – unless I count the doubles (20 =10 + 10) Hmm – into doubles? Cheers, Piccadilly! More about that solitary hare later.

We got out the tables and one clue yielded almost instantly and with 15625 at 1, we had either 53 or 59 at 2 since the digit sum of 2 had to be even. Simples! Using logic, we were able to fill a few more cells with the digit 1 and managed to suss that 27 had to be 125 but then seemed to be faced with a task that involved vast strings of possibilities for each solution. For example, 22 had to be a palindrome and we had sussed that 2 (59) was a factor of it so we had to do a Numpty troll though a long list of multiples of 59 that would give ?9?9?

12 and 9 had to intersect and it was that 7 that was the start of a long trail to the finish until the joyful moment when 23, a small digit sum that had to be a prime, and had to be greater than itself when reversed, produced the wonderful 11213 that confirmed all and roused a Eureka!

A mere ten pages of notes or so and just a few hours of mutters – “This isn’t maths, this is string-searching!” – and groans about numerical crosswords.  Well, many thanks, Piccadilly, and now I know that 5 is a factor of 15 etc. No, joking apart, how very clever to make all those statements in the clues correspond to the reality of simple arithmetic.

That elusive hare? I expected him to be gallivanting across the grid in alphanumeric style (8 1 18 5) but, in fact, it was a rather timid alphanumeric DOE crouching at the foot of the grid this time (4 15 5).

Posted in Solving Blogs | Tagged: , | Leave a Comment »