## Eleven Across by Elap

### Puzzle explanation

In this solution, the symbol ? denotes a digit.

5T + t = 1815: t must be a multiple of 5 and must start with 1 (because it has four digits). Since W and Y cannot start with a zero, t must be at least 1015 and so T must start with 1.

VV + t + x = 1423: x + t must be at least 1015 + 10 = 1025. V² is therefore 398 at most, putting V in the range 10 to 19. Rearranging the clue, t = 1423 – V² – x. The maximum value of t is therefore 1423 – 100 – 10 = 1313, putting t in the range 1015 to 1295.

T + 2W + a = 5220: T + a cannot be larger than 199 + 99 = 298, and so W must start with 2. This restricts the values of t to 1025, 1125 or 1225.

2S + t + xx = 4185: Rearranging, x² = 4185 – 2S – t. We know that t is 1025, 1125 or 1225. The minimum value of x² is therefore 4185 – 2*92 – 1225 = 2776, and the maximum value is 4185 – 2*10 – 1025 = 3140. This puts x in the range 53 to 56. We know that t + x² must be odd, ie, that x is even, and so we have the following two possibilities:
x = 54, 2S + t = 1269;   x = 56, 2S + t = 1049.
If 2S + t = 1049, t would have to be 1025 (not 1125 or 1225), making S = 12. This doesn’t work, because the second digit of S must match the second digit of t, and so 2S + t = 1269 (and x = 54) and t is 1125 or 1225, ie, 2S = 44 or 144, making S = 22 or S = 72. The second digit of t is therefore 2, giving t = 1225 and S = 22.

Returning to 5T + t = 1815, T = 118, and from VV + t + x = 1423, V² = 144, ie, V = 12. From L + V + Vx = 1184, L = 524, and from Sx + 2k = 1618, k = 215.

DD + DVV + Y = 8919: We know that V = 12, so we have D² + 144D + 5??4 = 8919. Let y (= 0 to 99) denote the two middle digits of Y. Then we have D² + 144D = 3915 – 10y, ie, D² + 144D = 2925 to 3915 in steps of 10. For D² + 144D to end in 5, the last digit of D must be 1 or 5. Only D = 21 produces a result in the required range, giving Y = 5454.

T + 2W + a = 5220: We know that T = 118 and that W ends in a 5, and so a is ?2.

K + VV + a = 2315: V is 12 and a is ?2, and so we have K + ?2 = 2171, giving K = 2??9.

H + 2W + q = 7189: We know that q is ??8 and W is 2??5, and so H must end in 1.

3G + j + 2k = 5195: We now know that j = ??19 and k = 215, and so we have 3G + ??19 = 4765, giving 3G = ?46, leading to G = 82, and hence j = 4519.

Q – H + 2j = 8938: Since j = 4519, we have H – Q = 100, and since H is ???1, Q is 1??1. q + v + 3w = 1514: We now know that q is ?18, v is 22? and w is 1?5, and so v must end in 1, giving v = 221. We now have q + 3w = 1293, ie, ?18 + 3w = 1293, and so 3w must end in 75, ie, w = 125. This gives q = 918.

Returning to T + 2W + a = 5220, we have W = (5220 – 118 – a) / 2 = 2551 – a/2. W therefore starts with 25, and we have W = 2525 and a = 52.

From K + VV + a = 2315, K = 2119.

Revisiting H + 2W + q = 7189, we have H = 1221, and from Q – H + 2j = 8938 again we have Q = 1121, supplying u = 1154.

F + SS + 2f = 4919: We know that S is 22 and F is 2??, and so we have 2?? + 2f = 4435. f must therefore start with 2, giving f = 2112 and F = 211.

N – Gk + h + k = 2315: We know that G = 82 and k = 215. After substituting these values, we have N + h = 19730, ie, ??9 + ??21? = 19730, giving N = 519 and h = 19211.

BS – G – g – u = 1857: We already have S = 22, G = 82 and u = 1154, and so 22B = g + 3093. Let the second digit of B be x, and the second digit of g be y. Then we have 22(2014 + 100x) = (10000x + 1000y + 215) + 3093. Rearranging, 7800x + 1000y = 41000. 7800x is therefore a multiple of 1000 and must therefore be 5 (remember that x and y have to be in the range 0 to 9), leading to y = 2. This gives B = 2514 and g = 52215, and E = 295 works itself out, thus completing the top right corner of the grid.

b + d + q + xx = 9147: b is ??1, d is ??8?, q is 918 and xx is 2916, and so we have b ( = ??1) + d ( = ??8?) = 5313. d therefore ends in 2 and b ends in 31.

C + F – 2e = 2118: F is 211 and so C = 1907 + 2e. Since C is 23??, e must start with 2, ie, e = 222. This gives C = 2351.

e + r – L = 1913: e is 222 and L is 524, giving r = 2215.

Returning to b + d + q + xx = 9147, the last three digits of d, q and xx are all known, and so b = 131, and this gives d = 5182.

X – A – F + f = 1919: Since A is 51?5, F is 211 and f is 2112, X must end in 3.

2s – E – P = 1919: E is 295 and s is ??13, and so P ends in 12.

H + R – J = 1920: H is 1221 and J is 1??2, and so R must end in 1, giving s = 2113.

G – F + Ss – p = 5203: p is the only unknown here, and a quick calculation reveals p = 41154.

From 2s – E – P = 1919 once more, we have P = 2012.

X – A – F + f = 1919: F and f are known and so we have X – A = 18. Since X is 5?43, A ends in 25, giving A = 5125 and X = 5143.

S – N – U + c = 1849: S is 22 and N is 519, and so we have c – U = 2346. c is 251? and U is 1?5, and so c ends in 1, leading to c = 2511 and U = 165.

L – H – JV + n = 1420: L, H and V are known and so we have n = 2117 + 12J. Let the second digit of J be x. Then we have 2117 + 12(1012 + 100x) = 20?61. Simplifying, this gives 14261 + 1200x = 20?61. x is therefore 5, leading to J = 1512 and n = 20261. This provides R = 2211, which is confirmed by H + R – J = 1920.

All the clues have now been solved. We have now to complete the grid so that all the rows and columns have something in common, but what is the link?

• Clue 1: The puzzle is called Eleven Across (but there isn’t an 11ac clue).
• Clue 2: There are lots of 1s in the values of the expressions in the clues.
• Clue 3: There are lots of 1s and 2s in the grid.

The predominance of 1s and 2s indicates that there is something special about the grid’s digits. Whenever there is an abundance of 1s and 2s, there is often a link to the position of letters in the alphabet. The first row in the grid is 512522514. Letting A=1, B=2, etc, this can be read as ELEVEN, ie, it reads as “eleven across”.

The values of the expressions in the clues are, in order,
4919 1857 1184 9147 2118 1920 8919 7189 4185 1618 5195 1420 1913 1815 1919 2315 1849 1423 8938 1919 1514 5220 2315 5203
and these can be decoded as
DIS REG ARD ING BAR ST HIS GRI DRE PR ESE NT SAC RO SS WO RDI NW HICH AIS ON EBT WO ETC
ie, “Disregarding bars this grid represents a crossword in which A is one B two etc”.

The rows and columns of the grid represent the following words (all of which are in Chambers Dictionary):
ELEVEN, WELLIE, URUBU, OLLAS, EXINES, TABULA, BULBAR, PEAVEY, ENCODED;
EVOVAE, MOTZA, YANKED, ERBIUM, VULVA, BALDLY, EVOKED, SUABLE, DESIRED.

The sum of the 44 grid entries is 192113, which can be interpreted as SUM.