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For more on marking an answer as the "Best Answer", please visit our FAQ.Finally finished the puzzle after having to restart because of an arithmetical mistake. I would think this would be rated as average difficulty for the Listener and I reckon there is only one grid pattern possible for grid 3 following all of the hints in the preamble. Definititely complete in the order grid 2, grid 3 and then grid 1. The hardest is grid 2 so once you've cracked that, you well on the way! Good luck - hope you get as much brain training from it as I did!
After a lot of wrong turnings I finally teased this one out. As far as I can see there are no ambiguous entries but.... there are numerous entries that have ambiguous clues. So beware! and assume nothing. There may be primes that are also palindromes, fibs that are also primes and so on. I personally figured out the grid pattern for 3 and filled most of it and then worked 1 and 2 in tandem. Good luck!
What most astonishes me about this crossword, and many of the other mathematical crosswords, is the brain of the setters and how they construct one of these grids so that there is a logical progression through the solving and there aren't any ambiguities to the final answer. This puzzle is such a web of cross-relations. Amazing!. Thank you, Oyler.
What most astonishes me about this crossword, and many of the other mathematical crosswords, is the brain of the setters and how they construct one of these grids so that there is a logical progression through the solving and there aren't any ambiguities to the final answer. This puzzle is such a web of cross-relations. Amazing!. Thank you, Oyler.
Agreed - guess we have to assume there is only one pure correct overall solution - though as I said above, I found another option very close to being spot on ... and within the confines of each specific grid (ie ignoring any relationship to others), there may well be several 'correct' possibilities.
Well done Oyler - though I might have preferred a slight bias towards pure deduction rather than this time round's over-emphasis on trial & error of possible combinations of solution.
Good fun nevertheless.
Well done Oyler - though I might have preferred a slight bias towards pure deduction rather than this time round's over-emphasis on trial & error of possible combinations of solution.
Good fun nevertheless.
I can't say I regard this as good fun. More like sheer slog with a lot of trial and error. Listener number puzzles should have a logical progression, and trial and error, though inevitable, should not dominate. While I was able to assign all of the clues for C, b and d to their respective grids from the start, and enter 2 digits in grid 2, from there on it was trial and error, with lots of error.
I've completed grids 2 and 3, only to find that both have squares for B, but neither has a square factor of D (I have the prime in grid 1). More troublesome is grid 1. I almost filled it before grids 2 & 3, then came unstuck, but after working on those I'm pretty sure my original entries for A, B and b in grid 1 are correct, but I haven't been able to get one or more of D, E and c and f to work out according to the clues.
Am I right in thinking that the entry for b in grid 1 has the same digit appearing 3 times?
I've completed grids 2 and 3, only to find that both have squares for B, but neither has a square factor of D (I have the prime in grid 1). More troublesome is grid 1. I almost filled it before grids 2 & 3, then came unstuck, but after working on those I'm pretty sure my original entries for A, B and b in grid 1 are correct, but I haven't been able to get one or more of D, E and c and f to work out according to the clues.
Am I right in thinking that the entry for b in grid 1 has the same digit appearing 3 times?
I think I've spotted the error with grid 1; that seems to work out now (with the 3 occurrences of one digit in b that I referred to earlier). But either grid 2 or grid 3 is wrong because neither has the square factor of D. I suspect it's grid 3 that's wrong, which I found the hardest anyway, unlike others.
Thanks for confirmation, Mysterons. I shall probably defer reconsideration of grid 3 until tomorrow, but I'm now beginning to think that the preamble may be right to suggest that solvers leave grid 3 until last (with 'Finally' in the last sentence of the frist paragraph). I started with grid 1, making quick progress until I reached an impasse, as I said before, with around 5 cells I couldn't fill; but now I've completed it I see that all but 4 cells in my original attempt were correct.
Well, I still can't get started correctly. I can see that where A=3G there can only be a limited number of possibilities for A but I can't make any of them work. Does A have 4 digits and if so are there only 4 possible answers? I think that I will stick to proper crosswords! No doubt I've missed something obvious.
Jamesah, one of the difficulties of this puzzle is that it's possible to fill ( or almost fill) a grid on one set of clues, only to find that it later turns out to be wrong. There are at least two solutions for grid 1, until you assign the clues correctly in the light of grids 2 and 3. 2 of the entries for A are divisible by 3 to give an entry for G. To help you decide which is correct it's worth studying the down clues that cross them.
The easiest grid to get started on is grid 1. With a little bit of experimentation all the cells for A, B and b can be filled, which is a very useful start.
The easiest grid to get started on is grid 1. With a little bit of experimentation all the cells for A, B and b can be filled, which is a very useful start.
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Jamesah, you have now been given the answer, so what follows may seem redundant, but it does show a route that succeeded for me; if it hadn't succeeded, the failure would have given me some useful information.
I started off with the assumption that A in grid 1 was either a cube or Fibonacci (not a logical deduction, but a working assumption to get me going). This narrowed the field down to 10 possibilities in theory, but all but 2 of those could be eliminated because of what the clue for b tells us (one of the few clues sets where it is possible to assign the clues to their respective grids at the beginning). Knowing that B was either prime or square it was possible to narrow the choices for that down to about 5 numbers. This entailed only 10 calculations, and one of them looked interesting when I considered the clues for G.
I started off with the assumption that A in grid 1 was either a cube or Fibonacci (not a logical deduction, but a working assumption to get me going). This narrowed the field down to 10 possibilities in theory, but all but 2 of those could be eliminated because of what the clue for b tells us (one of the few clues sets where it is possible to assign the clues to their respective grids at the beginning). Knowing that B was either prime or square it was possible to narrow the choices for that down to about 5 numbers. This entailed only 10 calculations, and one of them looked interesting when I considered the clues for G.
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