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!