Listen With Others

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 8^th 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=14, F=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).

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.