## The Domino Effect by Googly

### Detailed Solution

(1) 6a = DODO, 7d = LARD + RED – PP, 25a = LARD + RED – DODO, so 7d = 25a + 6a – PP, and must be 1•••. 2d = OO, so 4 ≤ O ≤ 9; 6a = ••1, fixing O and D, as well as 2d and 6a. O = 7
D = 3
(2) 23d = COAL + CARP – CARD, 33a = COAL + CARD – CARD – COP, 29a = COP, so 23d = 33a + 29a. Since 23d and 33a end in the same digit, 29a = •0. Hence 23d = 100&bull. 29a = 7CP, so it is 70 and 33a = 93&bull. Also, CP = 10, so C, P = 5, 2 or 10, 1. 11a = 7(C + P). 11a ≠ 49 since 2d = 49; P, C = 10, 1 and 11a = 77. 32d = A and 35a = PA, which fixes P, C and shows 32d = xx and 35a = xx0.P = 10
C = 1
(3) 28d = AA – DO = AA – 21 = ••0, so AA ends in 1 and A in 1 or 9. From 32d and 35a, the only possibility is 11, giving 35a = 110, 32d = 11 and 28d = 100. A = 11
(4) 23d is a multiple of 7, since the first term contains O and the others toether have a factor of P – D. Hence 23d = 1001 (not 1008 since the last digits in 33a must be a domino). This fixes [3 1].
(5) 30d = OR = 7R, 34a = C + ORA = 77R + 1. The only possibility is R = 8. Then 23d = 77L + 616 = 1001, giving L. R = 8
L = 5
(6) 25a shows E < 6, and so 7d = LARD + RED – PP = 1220 + 24E must be 1268 (E = 1, 25a = 927) or 1316 (E = 4, 25a = 975). 16d = LADDERED – BOD – PLOD + CELL = 189130 – 21B, if E = 4, which gives no valid values of B, so E = 2, fixing 25a and 7d. Then 16d = 46520 – 21B. 27d = 67B, so B < 15. Hence, to agree with a central 2 in 25a, B = 12, 13 or 14. Only 13 agrees with the 1 ending 27d. This fixes 16d, 27d, 14a and 12d. E = 2
B = 13
(7) AT SE corner, we must not repeat [1 0], so we can place [1 1] [0 0] [1 4] [2 6]. We cannot repear [4 1] so in NE corner we have [4 4] [1 2], with [• 6] in row 3 and [• 4] in row 4. In row 9 we have [6 1], while in row 8 we have [0 5] since we cannot repeat [0 0].
(8) 21a = 10 + 2U and must be greater than 66 (to avoid a partial domino). This gives U and 21a. U = 29
(9) 19d = FOLDED + LORD + COO = 630F + 889 = •••9. In column 7, the 1 must belong to [0 1] or [5 1] since other such dominoes have been located. Hence 20d ends 90 or 95. 20d = HOD – BEDE = 21H – 156, so H = 26 or 31. 1a = O + OH = 7(H + 1) and ends in 4, fixing H, as well as 1a, 20d and the position of [5 1]. H = 31
(10) 1d = DUE + ORE = 286, which fixes [2 2] [4 •] in row 1. 6d = SPAC + E = 110S + 2 = 47••, fixing S. This fixes [2 4] in row 4, and [• 3] in row 3. 13a = SEARCH + ORB = 235336, fixing [6 3] and showing row 3 must contain [6 •] to avoid duplicating [6 3]. S = 43
(11) 9d = LEARN + PARED + RAPS – CLOSE = 880N + 40110. It must have the form xy0••, where x = 7,8,9 and y = 0,4,5,6. There is only one possible value for N, fixing 9d, [6 0] in row 3. All 3s are now in the grid, so [3 3] is in row 5. N = 34
(12) 4d = DUCK – USED = 87(K - 86) gives K, and so 4d. In row 4, the 3 belongs to 0 or to a vertical domino, with 0, 2, 4 or 5. None of the latter set is possible since 17a is a multiple of 3. This fixes [3 0]. 3a = EGG = 37•, 57• or 67•, since 1a ends with a 4 of a horizontal domino. This fixes G and so 3a, as well as [4 5]. 17a = 3T = •3 fixes T as 21 since 11 and 31 are already determined (A and H). Column 1 contains [6 •]. K = 95
G = 17
T = 21
(13) 7d = 7204. Row 3 fixes [5 3], so 5d must be 87, 88 or 89. Since it is EX = 2X, this fixes X. 22d = HEEL + DADDY – ADDED = 26 + 297Y = 4••9, 5••9, 6••9, since there is [6 •] in column 1. This fixes Y, 22d and [6 5] [6 6] in column 1. 15a = 89. We have [2 3] in row 3, [3 4] in column 3 and [0 4] in column 4. The only remaining 4 is [6 4] in row 5. X = 44
Y = 19
(14) 19d = FOLDED + LORD + COO = 630F + 889 = 6••9, giving F. [5 5] is in column 6. 26a = VOLVA + PLATE = 385VV + 23100 gives V. Then [0 2] is in row 7 and [0 1] is in row 6. F = 9
V = 41
(15) The only remaining domino is [5 2], which must fit in 24a or 15d, with the common digit 2 or 5 and the remaining digit 7, 8 or 9. The only possibility is 25/875. Q = 25
I = 35
(16) Factorise the unclued entries:
8a = 87978 = 2.3.11.31.43 = EDAHS → SHADE
31a = 79170 = 2.3.5.7.13.29 = EDLOBU → DOUBLE
18a = 476 = 2.2.7.17 = 7.34.2 = ONE
and so the domino [1 1] must be shaded. (Specifying the layout of all the dominoes was optional.)