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Have Fun, It’s Wordy by Kea

Posted by shirleycurran on 8 March 2019

I suspect Kea is being somewhat cynical in his title, knowing that a number of regular solvers consider these three-monthly numerical crosswords to be the reverse of fun. We puzzled over the rather obscure (at least we thought so) pre-ramble for some time without understanding what we were entering at A and n. The other Numpty then decided that the only possibilities for A were 11 or 12, which would give us 53 or 82 at n, and he disappeared to create a searchable programme in BASIC to give us 10 to 999 with the total number of letters in each case, and the alpha-numeric equivalents of distinct letter totals. We still bungled along for a few hours, repeatedly confusing which numbers went into which cells (and not finding a trace of alcohol in the grid – Oh dear Kea!) – but we got there.

A friend commented that he had been able to work out a logical path and he very kindly sent it to me so enough of our computer-aided bungling. Here’s how we should have done it:

Have Fun, It’s Wordy by Kea

Note that if you know an entry, you can immediately compute its value and the length.  Whether it’s fun or not is debatable😊.                                                                                                                                                                                   

The only 2-digit numbers whose spellings have 6 letters are 11, 12, 20, 30, 80 and 90. A cannot end in 0 so is either 11 or 12.  If A = 12, then n = T+W+E+L+V = 82 and G = E+I+G+H+T+Y+W+O = 112 which is too big.  (That is the last time I will show the letter addition!)  So A = 11, n = 53, G = 91, d = 88, F = 74.

h = 4xy.  Any possibility other than 400 would be spelled FOUR HUNDRED AND <something> which is more than 14 letters.  So h = 400.

M is then spelled <h> HUNDRED AND <y> where <y> is from 10, 20, 30, …  and <h> and <y> use eight letters.  H+U+N+D+R+E = 71.  The additional letters in <h> and <y> must sum to 20.  The only possibility that works is M = 310.   (maybe you can figure out a better explanantion).

p is 31, 32, or 36; D is then 114, 118 or 123.  Only p = 32, D = 118 work with d = 88.  From D, M-H=122, so H = 188, E = 147, f = 178.

g = 41x and only g = 417 has 23 letters,

K = 5×7 and must be 527, 537, 587 or 597 to have 25 letters.  Those numbers have values 195, 172, 179 and 172 respectively.  d+G = 179, so K = 587.

k = 18x and is a multiple of A=11 from clue C, so k = 187. 

c = x17 and must be 117, 217 or 617 to have 22 letters, with respective values 147, 170 or 165.   Since E = 147, c = 117.

N = 7×2 and must be 742, 752 or 762 to have 23 letters, with respective values 201, 210 or 228.  Since B ends in 1, N = 742, B = 201, e = 129.   B-b = d+G = 179, so b=22.

C = 1×2 and has 23 letters; only C = 172 works.  a = 171.

m = 81x and must be 815 or 816 to have 22 letters, with respective values 113 and 150.  C – b =150, so m = 816.

Finally P = 6x and must be 63, 67 or 68 to have 10 letters, with respective values 128, 138 and 117. c = 117, so P = 68.

‘nuf said, thank you, Kea.

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