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Listener 4151: Number or Nummer by Ruslan

Posted by erwinch on 9 September 2011

Ruslan’s third Listener but this was not so much a numerical puzzle, as were the previous two, but a riddle in the form of a preamble.  For a long time I could not make any sense of it at all, with good reason as it turned out, but finally managed to struggle to a finish.
The key for starting here was a quartet of four clues with entries in the SE corner of each grid:
24ac  D^2 (3)
19dn  A^2 (3)
15dn  BD^2 (4)
21ac  ABI (4)
We were not told in the preamble but soon learnt that checked cells could sometimes contain different digits which had the same first letter.  So, for any one cell we had to consider across, down, German or English – I found it all terribly confusing, worse than working in three dimensions.  Anyway, D and A must be greater than 7 for the square to be three digits and the squares crossed at the final digit.  So, D^2 and A^2 were found in {121, 361, 841} or {169, 289, 529}.  I went wrong initially by assuming that BD^2 (15dn) meant B squared times D squared – there was a fit with B = 2 but ABI was then too small.  The final digit of BD^2 crossed the middle digit of D^2 and nine possible fits were found:
Now looking at ABI (21ac), for any one value of BD^2 there were two possible values of A and seven of I.  ABI also had to fit with crossing BD^2 and A^2.  Two fits were found: {ABI, BD^2} = {2233, 2527} or {5423, 6137}.  So, B = 7 or 17, A = 11, D = 19 and I = 29.  With the T at cell 21 in the left-hand grid given, we could now make our first confirmed entries:
21ac ABI = 2233(TTTT or ZZDD) and 5423(FFTT or FVZD): TTTT is the only fit with the left-hand grid:
21ac ABI = 2233(TTTT)
19dn A^2 = 121(OTO or EZE) but must be OTO to fit with ABI
24ac D^2 = 361(TSO or DSE) but must be TSO to fit with A^2
15dn BD^2 = 2527(TFTS or ZFZS) but must be TFTS to fit with ABI
For the right-hand grid:
21ac ABI = 5423(FFTT or FVZD)
19dn A^2 = 121(OTO or EZE) but must be OTO to fit with ABI(FFTT)
24ac D^2 = 361(TSO or DSE) but must be TSO to fit with A^2
15dn BD^2 = 6137(SOTS or SEDS) but must be SOTS to fit with ABI
So, our first entries were all in English:
As far as the numbers went, the remainder was plain sailing starting with 1ac (lg = left-hand and rg = right-hand grid):
1ac  (Z + I)^2 – AZ – F – T (3)
The maximum available values of F and T for both grids are 23 and 19 giving F + T = 42.
If Z = 2 then 1ac is greater than equal 31^2 – 64 = 961
If Z = 3 then 1ac is greater than equal 32^2 – 75 = 949
If Z = 5 then 1ac is too big
So the first digit of 1ac and 1dn is 9(N) and Z = 2 or 3
14ac  (A + Z)(A + B + I + Z) (3)
(lg) 14ac = ?T? so must be English
If Z = 2 then 14ac = 13 × 49 = 637
If Z = 3 then 14ac = 14 × 50 = 700 so Z = 2 and 14ac = 637(STS)
(rg) 14ac = ?S? English or German
If Z = 2 then 14ac = 13 × 59 = 767
If Z = 3 then 14ac = 14 × 60 = 840 so Z = 2 and 14ac = 767(SSS)
22dn  B + A + L + Z (2)
(lg) 22dn = TT = 20 + L so L = 3 or 13
(rg) 22dn = FT = 30 + L so L = 13 or 23
18ac  I – B – L + ZA (2)
(lg) 18ac = FO = 44 – L = 41 or 31 so L = 3
(rg) 18ac = OO = 34 – L = 21 or 11 so L = 23
Things appeared to be going very well until I looked at the NW corner of the grids:
8ac  Z + ID^2 – (B + L)(I – D + AD) (4)
(lg) 8ac = 8281(ETEO or AZAE)
(rg) 8ac = 1711(OSOO or ESEE)
1dn  I(A + D) + B(A – 2Z) (3)
(lg) 1dn = 919(NON or NEN) but had to be NEN (German) to fit with 8ac ETEO
(rg) 1dn = 989(NEN or NAN) but again had to be NEN to fit with 8ac ESEE (German)
But this would mean that we had German entries in both grids, which I had not been expecting from the preamble:
Solvers must identify the English transcriptions of the answers entered using German in one of the grids …
With hindsight I can see that this is ambiguous but at the time I naturally assumed that German entries would appear in one grid only leaving the other as the intended solution.  How could there possibly be a single solution with German entries in both grids?  At this point I came to a standstill, agonising over the whole thing for ages.  I regret not pushing on immediately to see how it came out.  Eventually of course, I did go on and all became clear.  Mostly it was just straightforward substitutions required to complete the grids so it is unnecessary to give any more working but here are the full assignments for the two solutions:
And here is the full solution:
Some entries such as 6dn (765 777) and 10ac (65 75) in both grids were the same in either German or English but those that could only be German gave in clue order:
(lg) FEE/ZNEZA NEN/SANSA  511/29128 919/78968  FOO/TNOTE NON/SENSE
(rg) ESEE EZNNF  1711 12995  OSOO OTNNF
I was annoyed at having been tricked by Ruslan but can see the funny side now.  It was all nonsense – there was no local solver in Germany and there would have been no solution without the German entries.  It was a bit like those weak dramatic plot-lines where it all turns out to have been a dream.  One thing that was not weak was the construction, which appeared to me to be mind-bending in its complexity.  I feel as though we have been given a glimpse of life at Bletchley Park – I doubt that I would have been of much use there but am certain that Ruslan would have been and indeed perhaps was a valuable asset.  Anyway, this was sometimes frustrating but still tremendous fun – many thanks R!
Will Elap be able to continue his impressive run and produce a numerical Listener for a ninth consecutive year?  We will find out in November.

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