Solution to the Dogsmead Cross-Number Puzzle Let f be the age of Farmer Dunk, m the age of Mary, w Dunk's walking speed in mph, x the length of Dogsmead (yds), y the breadth of Dogsmead (yds), v the value of Dogsmead in shillings per acre, k the solution to 9 Down. Then, among others, we have: General: Y. xy/1210 Across: 1. xy 6. |x - y| 7. xyk/1210 10. f 14. 2(x + y) 15. w-cubed 16. w-cubed - k Down: 1. v 3. m 4. xyv/(4840 * 20) 6. 2m + 1 7. y-squared 8. (x + y)/11w 9. k 10. kf Abbreviating Across and Down to A and D respectively, we note: (a) 11A starts with the digit 1. So k ends with 1. (b) 15A is a two-digit cube, so it must be 27 or 64. If it is 64, then w-cubed - k must end in 3. But then 7D ends in 3, which is impossible since 7D is a square and no square ends in 3. So 15A is 27, and w is 3. (c) 16A has two digits. So w-cubed - k >= 10. So k <= w-cubed - 10 = 17. But k (= 9D) has two digits, and from (a), ends in 1. So k is 11 and 16A is 16. (d) kf (= 10D) ends in 2. But k = 11. So f ends in 2. But then, 10A ends in 2, which means that 8D also ends in 2. Now 8A starts with 1 (since it is a 4-digit date before the mid-20th century.) So 8D starts with 1. So 8D is 12. But then (x + y)/33 = 12. So x + y = 396, and 14A is 792. (e) Three of the four digits in the second column total 10, and 12D ends in 9. So 12D is 19 (and, though this will be found independently, the second digit of 1A is 8.) (f) From general clue Y, 11|xy. So 11|x or 11|y or both. ("is a factor of") From 8D or (d), 11|(x + y). So 11|x AND 11|y. From general clue Y, 5|xy. So 5|x or 5|y or both. From 7D, y-squared ends in 6. So y ends in 4 or 6. So y is even, and 5 does not divide y. So 5|x, and since x + y is even, x is even. In summary, 2|x, 5|x and 11|x, that is, 110|x. But x + y = 396. So x is 110 or 220 or 330. But from 6A, |x - y| has only two digits. So x = 220 and y = 176. Whereupon, 1A = xy = 38720, 6A = |x - y| = 44, 7A = xyk/1210 = 352, 7D = y-squared = 30976, and Y = xy/1210 = 32. (g) And now we have found that 6D is 45, so that 3D = m = 22. (h) Now 4D is 2v/5. And, since 1D (= v) begins in 3, 300 <= v < 400. So 120 <= 2v/5 < 160. But 4D ends in 42, so it must be 142. Whereupon, 1D = v = 355. (i) We have seen from (f) that general clue Y is 32, and this appears by chance somewhere in the puzzle. The remaining candidate places are 5A and 10A. But f, the age of Farmer Dunk, must exceed the age of his first-born. So 10A must exceed 6D which is 45. So 5A must be 32. (j) This implies that 2D is a 4-digit square beginning with 73. The only square between 7300 and 7399 is 7396, or 86-squared. (This surely rules out the possibility that Mrs D is the mother of Mary; but nobody ever said that Mrs D was even a relative of the Dunks!) (k) From the first three digits of 11A, Mary was born in the 1910's. And since she is 22, the "present" year must be in the range 1932 to 1941. From 8A, the Dunks first occupied little P in 1610. So the first digit of 13D is 3. So 11A is 1913, and the "present" year is 1935. 13D is 325. (Yes, we have made some assumptions about birthdays; also about the exact interpretation of the clues. The latter could change the final digit of 13D, but there is no way to confirm what the author intended.) (l) It remains to determine 10A (= f) and 10D (= 11f). f ends in 2, and, because of 3D, surely f > 52. This leaves only 62 and 72 as possibilities, since 11f has three digits and must begin with the same digit as f. If f is 62, kf is 682; if f is 72, kf is 792. Only the latter causes general clue X to be obeyed. And this completes the puzzle!