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4062: Elap’s Numbers and Letters (or 17 25 2 11)

Posted by Dave Hennings on 18 December 2009

I suppose I always approach numericals with a certain trepidation, although they have normally been relatively easy to solve. The exception for me was 4010, Euclid’s Algorithm by Aedites which consisted of a cicular grid and 145 clues! It coincided with a time when I was living in a hotel room for six weeks, and solving the Listener outside my normally cosy environment was difficult. I had already failed miserably very early in the year, so an all-correct was not on the cards. I hoped that Elap’s puzzle would not be too difficult … it would be unfair, I think, to be tripped up by a numerical, especially the last of the year (although one all-correct last year was caught out by Aedites).

It looked an interesting puzzle: no bars, and either one or two numbers in each square (1-25), so a bit of novelty there. Changes would be required after the grid was complete, with some sort of message appearing, and ambiguities to resolve. This last bit was the only thing that really worried me.

I found the way in fairly quickly: 18ac required K to be 4 or 9, and U 9, 16, 25 or 36, and 2ac being 4 digits required UU to be at least 920, ie 36. I am sure others will go into the full analysis of the solving method, so I will leave that to them. Suffice it to say that after I completed most of the grid I could sort the values of each letter and the first fourteen appeared as KEY SUBMIT XWORD, and replacing the number in each square resulted in a grid full of 7-letter across and down entries.

Identifying the likely words also meant that I didn’t have to solve the last few mathematical steps, which looked horribly like requiring a bit of quadratic equation solving.

I looked at the grid and felt pretty pleased with myself. There was just one problem … I hadn’t spotted any anomolies, and I suddenly wondered whether there were two solutions that were possible, which my corner-cutting had hidden from me! Perhaps one or two words had alternatives, differing by just one square. There was only one course of action that I could take … solve the whole bloody thing again, and this time right to the end. I don’t think I needed quadratics in my second solve, and managed to get C and N with just a bit of substitution between 8dn and 16dn. In the end, I only had one ambiguity in square 6 of the bottom row which could hold either 2 or 24. My original solution also proved to be correct.

Another excellent puzzle from Elap, although he will have to go a long way to give me more enjoyment than 3957, Eleven Across, 3879, Digital Diversion, or (especially) 3853, Sevens and Threes.

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