This is IOA’s tenth Listener Crossword (or crossnumber – it seems that most of them were numerical puzzles).
Our numerical puzzle blogger, Gill, solved this one and these were her comments to IOA:
Everything was going so well. 1ac, 15ac and the redundant 2dn gave me 195, 351 and 546 respectively, and I was off to a flying start. Rather more work was required to arrive at 961 for 16ac and then I reached a 3-branched fork for 10dn: 289, 529 or 729. I embarked on 289, but by this time the birds were tuning up for their dawn chorus and I was in danger of making an error and undoing hours of work. So I called it a night (or rather a morning) and went off for some zzzz’s.
I hoped that the next morning would bring a pdm but it was clear that persistence and careful recording of paths and their dead ends were needed. Unfortunately examining all options for 10dn did not bring a fast resolution but branched out into 2 possible grids. So I turned my attention to the SW-NE diagonal. Only partly filled, it fell in the range 35-60 and by creating a table of products of possible values and multipliers, I narrowed 14ac down to 284 287 248. Meanwhile 9ac was emerging as 243 and so I took a deep breath and plumped for 287 for 14ac, the only product which gave a SW-NE diagonal in the range ie 41.
I had now solved all the across clues, and of my 2 possible grids, the first one in which 10dn was 529 proved to be the viable one. 7dn proved to be 289 but the NW-SE diagonal didn’t yield a quick win. In fact I felt I had far too few down clues for the number of blanks cells left. There was nothing for it but to examine 2 possible final digits for 4dn: 7 or 9. I started with 7 and everything was going swimmingly until I came to fill in the centre right square – clashes everywhere!
That just left 9 for the final digit of 4dn. By this time I could narrow down the sum of the NW-SE diagonal to between 48 and 55 and this gave 612 for 13dn, 479 for 4dn and 361 for 3dn. This gave 51 for the diagonal – phew! It worked!
If some wiseacre solves this by asking AI, I don’t want to hear about it!
Thank you IOA for a thorough brain workout. Credit to you for a cleverly designed sudoku with minimal rubric and clues. Sudokus claim to stave off dementia – well, I should now be good for a decade or two yet. And you for even longer, for the ingenuity required to set such puzzles!
Here’s Gill’s solution path:1. 1ac is 2dn – 15ac but
2. 15ac = 9/5 x 1ac
3. So 2dn = 14/5 x 1ac; 1ac ends in 5; 2 dn begins with 5
4. 1ac = 195; 2dn = 546; 15ac = 351
5. 10dn ends in 1 4 5 6 9; 11dn ends in 2 4 6 8
6. 16ac is a square: 169 529 625 or 961; 11dn ends in 2 or 6
7. Therefore 5ac (not containing a 4) ends in 1 or 3 (29 possibilities)
8. 8dn ends in 3 or 9, starts 1 or 3 and does not contain a 2 (10 possibilities)
9. Therefore 6dn ends in 6 or 8: the 9 candidates are 278 298 326 346 358 386
698 718 758 – 778 (389×2) eliminated
10. 3dn ends in 1 4 9 and does not contain a 5 or start with a 1 or 9; candidates
are 289 324 361 729 784
11. 4dn does not contain 1 or begin with 5 or 9. 4 is not the middle digit. 34
candidates!
12. 9ac ends in 1 3 7 9; possibilities are 243 or 729
a. Let 9ac = 243
13. 8dn = 173 193; 6dn = 346 386
b. Let 9ac = 729
14. 8dn = 139; 6dn = 278
15. 12dn begins/ends in 1: 127 163 167 173 271 281 421 461 521 541 571 641
761 821
16. 13dn = 21x 41x 61x 71x 81x 91x where x is not 1 3 5: possibilities are 216
217 416 418 612 714 812 814 816 817 819 912 918
Three options for 6dn: 278 346 386
a. Let 16ac = 169
17. 12dn begins with 1: 127 163 167 173
18. 10dn ends in 1: 841
19. 6dn = 346 386 or 278 – No solution for 6dn
b. Let 16ac = 529
20. 10dn ends in 5: 625
21. Clash of 2s and no solution for 6dn
Another solver kindly sent me his solution path: if, like me, you are a total Numpty with numericals, this ‘easy’ solve might be an eye-opener.
Did it by writing code; outline of solution; at each step make sure there are no repeated digits or zeros in any row, column or box
Find all triplets 1a, 15a 2d such that 2d = 1a+15a, 15a = 1.8 * 1a, 2d is 3 digit, and 1a and 2d agree but repeat no digits other than their intersection:
There’s only 1 such triplet: 195, 351, 546
Next deal with 5a, 11d, 10d 14a and 16a to get most of box 7
Then do 8d, 6d, 7d, 9a
Next 3d, 4d
Generate candidates for the unclued cells in the SW-NE diagonal and determine 14a = 287
At this point you can do some normal sudoku logic to infer several of the unclued cells and determine that 12d = 163
Get candidates for 13d and the unclued cells in the other diagonal.
Sudoku rules force 3d and 4d
After that it’s just a normal, easy sudoku.
I often give the numerical puzzles a miss, but this one, with its straightforward rules and a minimal set of clues, had an instant appeal. After reading all the clues I concluded that a related set of four of them (1ac, 15ac, 2dn and 11dn) was key to making a start, and the rest of the puzzle would need a reference table of three-digit squares and primes and the three-digit doubles of some of those primes. All numbers having any zeroes or repeated digits would be excluded from these lists.
The two ‘multiples’ were a clever touch: I took them at first to be a way of confirming one’s entries towards the end, but (unless I’m mistaken) they were actually needed in order to resolve not only the two diagonals involved but also the NE and SE corner boxes. I was impressed by the fact that a unique solution could be forced from the minimal information given in the instructions and clues.
Thanks to the setter, and to Gill for having the patience to document in detail all of her workings in arriving at the unique solution. I’m glad my completed grid looks exactly like hers!
Not a crossword. Boo
I haven’t attempted a numerical Listener before but I found this quite accessible and enjoyable. Looking forward to the next one now. It was nice that those first three clues gave a few digits that could be filled in early on – I might have given up if it had taken a long time to get anything at all into the grid.
Excellent, elegant, puzzle – many thanks, IOA. Took much enjoyable time to get to the end, which I (eventually) did by the logical path with the diagonal sums the last bit. I also found it useful to draw up lists of the relevant squares (13 in all) and primes (89).
Thanks also to blogger(s).
Re Aphld at #2, quarterly numericals go back a very long way in the Listener and are popular (similar or sometimes more entries than average). Technically not a crossword of course.