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Listener 4138: The X-factor by IOA

Posted by erwinch on 10 June 2011

A third Listener from IOA whose first eleven years ago was not numerical.  It was entitled IEEOO and only occurrences of the five vowels in answers were entered into the grid – hence the real title: Listener Crossword.  Also, I thought at the time that this would probably account for the pseudonym IOA but although you can form any number of names from three vowels I did not know of any candidates that fitted.  It is a bit different now though and we can probably rule out Vicky Pollard but I wonder, might IOA be Simon Shaw?
 
But back to the present and since they had double clues we knew immediately that A, R and V would be 1, 5 and 13 in some order.
 
From K. = H we know that K must be a two-digit entry (4, 12, 17 or 23) except that K = 23 is too big for R. = OKKK.  K = 17 will also be too big if O is greater than 2.
 
From W. = (WW – XXXXX)O we see that XXXXX < 676 (26 × 26) so X = 2 or 3
 
From U. = UUUU – UUU + UU + K only two values of U. are found that fit at the corresponding grid entry: U = 7 or 9
If U = 7 then U. = 2107 + K = 2111, 2119 or 2124
If U = 9 then U. = 5913 + K = 5917, 5925 or 5930
 
L. = UV so V cannot equal 1: V = 5 or 13
 
From V. = OOOOU – O – U
If O = 2 then V. = 103 or 133
If O = 3 then V. = 557 or 717
If O = 4 then V. = 1781 or 2291
If O = 6 then V. = 9057 after which V. is too big (greater than four digits) for further values of O
 
But O. = QU so O. cannot be 2dn (four digits): O = 3, 4 or 6
 
From V. = (VXX + A)VXXX with V = 5 or 13; X = 2 or 3; A = 1, 5 or 13
If V = 5 then 5ac or 5dn will end with zero and 15ac or 7dn will begin with zero (prohibited) so V = 13
If X = 2 then V. = 5512 or 5928
If X = 3 then V. is too big: X = 2, 13ac = 557, O = 3, U = 7, L. = 91, L = 4, 12, 17 or 23; A = 1 or 5; 13dn = 5512 or 5928 and K = 4 or 12
 
Now looking at R. = OKKK: {R., K} = {192, 4} or {5184, 12}
From R. = HHH + XXXXX = HHH + 32 knowing that H > 9 this must have four digits so the R. above (1dn or 5ac) = 192; H = 11 or 12 and K = 4, 7dn = 2111
 
R. = 1ac (1??1) or 5dn (1???) = HHH + 32 = 1363 (H = 11) or 1760 (H = 12) but cannot end with zero so 5dn = 1363, 5ac = 192, R = 5, H = 11,  A = 1 and 13dn (V.) = 5512
 
So, the assignments to date are as follows:
 
 
 
A. = (D + D + V)DD = 900 (D = 6) then 1856 (D = 8) so this must be the four digit value 1ac (???1) and D < 16
Only D = 9 fits so 1ac = 2511
 
1dn (A.) = (B + U)(B + U) + BU = 2??
{1dn, B} = {211, 6} or {281, 8}
 
2dn (X.) = ((B + U)(B + U) + BU)S = 5??5
If B = 6 then 2dn = 211S (S = 25)
If B = 8 then 2dn = 281S (no fit for S)
B = 6, 2dn = 5275 and S = 25
 
9ac (D.) = DDKK + S = 1321
 
Y. = BBBKKR – BK = 17256 so Y = 10 or 19
 
11dn (H.) = WW – HH + HH = WW – 113 = 5?? (Y = 10) or ??7 (Y = 19)
{11dn, W, Y} = {563, 26, 10} or {287, 20, 19}
 
W. = (WW – XXXXX)O = 1932 (W = 26) or 1104 (W = 20) but 1104 does not fit at 20dn so W = 26, 26ac = 1932, Y = 10 and 11dn (H.) = 563 
 
N. = ON = 3N so N must be a two-digit entry and only N = 17 fits so 17ac (N.) = 51
 
We now have values assigned to fourteen letters and the grid looks like this:
 
 
 
I did not attempt to deduce the instruction at this point but with hindsight I see that I could have found the probable value of E (= 21).
 
We can see from 3dn (O.) = QU = 7Q = 1?2 that Q = 16 is the only fit giving 3dn = 112
Z. = HH = 121 and since all answers are different Z = 8
M. = BBBB + RRRR – AAAA – OOOO = 1839, which can only fit in the grid at 22ac and also places 91 (L.): M = 22, L = 23
 
And so it went until just one letter and one number remained unassigned: J = 12
 
This gave the completed grid:
 
 
 
I made a meal of interpreting the bottom row but the instruction was soon found:
 
Shade evens but rub out all of odds – ergo X
 
Giving the solution:
 
 
 
So, a straightforward puzzle, which required just two sides of A4 for working using a pencil and calculator only.  My enthusiasm for the Listener in general has waned somewhat since it peaked during the nineties – I blame the Internet – but it remains equally as strong when it comes to the numerical puzzles.  IOA certainly had the harder role as setter but I enjoyed every minute of solving this – thank you.
 
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