Listener No 4425: Clue-by-four by gwizardry
Posted by Dave Hennings on 9 December 2016
I needed to revisit our blogs for Listener no 4242 (Killer Queen) to remind myself of gwizardry’s first puzzle which combined mathematics and playing cards. That one was relatively tricky, but this week’s had a straightforward mathematical theme based on Lagrange’s Four-Square Theorem which he proved in 1770. We weren’t given bars or clue numbers in the grid, nor clue answer lengths.
At first, I thought this would be like the last mathematical (Nod’s Analogy) where we had to solve all the clues before we could start filling the grid. It soon became clear that this was not the case.
All clues ended with either D or T, so one was 0, the other 1. From a number of clues (eg 8ac RTTD) T was greater than D, therefore T was 1 and D was 0.
After that, the way in seemed to be from 1ac (GGDD), 37ac (GGGD and symmetrically placed) and 2dn (GXDD). Trying values for G (> X):
– G = 3: 1ac = 18, 2dn = 9 + X²: no X gives 2dn as 8•
– G = 4: 1ac = 32, 2dn = 16 + X²: X = 2 or 3
– G = 5: 1ac = 50, 2dn begins 0
– G = 6: 1ac = 72, 2dn = 36 + X²: no X<G gives 2dn as 2••
– G = 7: 1ac = 98, 2dn = 49 + X², X = 6, 2dn = 85, but 37ac = 147 and 1ac only 2 digits
– G = 8: 1ac = 128, 2dn = 64 + X², no X<G gives 2dn as 2••
Thus G = 4 and X = 2 or 3. Then 33dn (GDDD) = 16, 33ac ( RGXD) = 12• or 13•, with R =10 or 11. R = 11 gives too big a value for 33ac, therefore R = 10, and 8ac = 102, 2dn (GXDD) = 20 and X = 2. From there, the NW and SE corners began to fill fairly quickly with the odd foray into the other corners and consolidating in the middle.
I’m afraid that’s as far as my detail goes this week. It didn’t take long to find the thematic word QUAD (after trying FOUR) that went below the grid: Q²+U²+A²+D² = 24²+20²+12²+0² = 1120. The full list of letter equivalences was:
Yet again, we had a mathematical puzzle with a novel clueing mechanism. Not too long a solve, but good fun. Thanks, gwizardry.