No badger can guess a conundrum.

In reverse alphabetical order.

Chris Adams sent me the guided solution below. Throughout, L is the length, B the breadth, and all units are in yards/square yards unless otherwise noted.

1. Both 8 and 11 Across are years before 1939; since both are four digits, both must begin with a 1.
2. The only two-digit cubes are 27 and 64, so 15 Across must be one of these numbers, and hence Father Dunk’s walking speed in miles per hour is either three or four. Let S denote his speed in miles per hour; then 88S/3 is his speed in yards per minute. And thus the time it takes him to walk the perimeter is 2(L+B)/(88S/3) = 3(L+B)/(44S), and so 4/3 this time (8 down) is (L+B)/(11S), and this is between 10 and 19 inclusive.
3. Now, the perimeter is less than 1000 (since 14 Across is a three-digit number), so L+B, which is one half the perimeter, is less than 500. So L+B is equal to 11 times S times some number between 10 and 19, and this is lower than 500. Regardless of the value of S, 16 to 19 give us a value greater than 500, so the time is between 10 and 15 minutes.
4. Look at 9 Down, 10 Across, and 10 Down. Since 9 Down ends in a 1, and 9 Down times 10 Across equals 10 Down, this means that 10 Across and 10 Down end in the same number. This also means that 8 Down ends with the same number that 15 Across begins with. And 15 Across begins with either 2 or 6, while 8 Down ends with 0, 1, 2, 3, 4, or 5. Hence that number must be a 2, so it takes 12 minutes, and the speed is 3 mph, which yields 27 as a cube.
5. We know the speed and the time it takes to go 4/3 of the way around the perimeter, so we can solve for the perimeter and get that the perimeter is 792 yards. Alternatively, one can get that from L+B = 11×3×12, and P=2(L+B).
6. Now, 12 Down is the sum of the digits in 10 Down, plus 1. This must be 19, as any other two-digit number ending in 9 is greater than any possible sum of the digits. Hence 10 Down must be 792, 882, or 972, making 10 Across 72, 82, or 92, respectively. Only in the first case does 10 Across divide 10 Down, since 792 equals 72×11. Hence 10 Down is 792, 10 Across is 72, and 9 Down is 11. This also makes 16 Across 16.
7. 10 Down is a square and ends in 76, so B ends in either a 4 or a 6. Also, since L-B is less than 100, and L+B equals 396, B must be between 150 and 190, so there are eight possibilities for B. Of these, only 174 and 176 end in 76. However, 174 ends in 276, meaning Mary was born in the 1200s, which is plainly false (her father is only 72). Hence 76, which ends in 976, is the correct answer.
8. The breadth B is 176, so the length L is 220, and this makes the area (1 Across) 38720. It also makes the number of roods 32, so 7 Across is 384. Also, 6 Across is 44, and so 6 Down is 48, which is Ted’s age. In six years, he’ll be 54, which is twice Mary’s age then. So she’ll be 27 then, which makes her 21 (3 Down) now. It also means she was born in 1918 (11 Across).
9. Putting 8 Across and 13 Down together, we can fill in the blanks and show that the family acquired the property in 1110, so they’ve owned it for 829 years.
10. 2 Down is a square number ending in 1, so the original number ends in 1 or 9. It is also in the 7000s, and since 80 squared is 6400 and 90 squared is 8100, this makes Father Dunk’s mother-in-law 81 or 89. 81 doesn’t work, but 89 squared is 7921, and so the mother-in-law is 89.
11. 1 Down and 4 Down are still open, but we know that 1 Down is 5/8 (or 20/32, since there are 32 roods) of 4 Down. And 1 Down is between 300 and 400, so 4 Down must be between 480 and 640. Since 4 Down ends in 44, the only choice is 544, which makes 1 Down 340.

This puzzle inspired an even harder one by CUNY math professor William Sit.

1. Kh3! Kf4 2. Rf6#.

From The Tablet, 1920.

Square both sides:

Now we can substitute the original equation, giving x + 2 = 4 and x = 2.

1. Re1! If Black takes the rook then 2. Bc3 is mate; if he tries to flee then the queen finishes him from e2 or d1.

From Deutsches Wochenschach, July 22, 1917.

Imagine two clocks, one that runs normally and one that runs at 12 times normal speed, so that the hour hand of the fast clock always matches the minute hand of the normal one. Our short-handed clock will give an ambiguous reading every time these two imaginary clocks show the same time, provided we omit cases where the two hands coincide.

During the period from just after noon to midnight, the minute hand of the fast clock revolves 144 times and the hour hand of the normal clock revolves once. So the two clocks will show the same time 143 times. During that period the two hands of the accurate clock will coincide 11 times, so between noon and midnight we’ll be unable to tell the time 143 – 11 = 132 times.

The same reasoning obtains for the period from midnight to noon, so the answer is 264.

From “A Clock Puzzle,” by Andy Latto, in Puzzler’s Tribute, edited by David Wolfe and Tom Rodgers, 2002.

It remains unchanged. A floating body displaces its own weight in water, so dumping the ice has no net effect: instead of one floating body we now have two with the same total weight. This remains true as the block melts; in effect the ice occupies a “hole” in the water exactly equal to its own melted volume.

The fives and the tens. Without these, only a four-card run (A-K-Q-J, 9-8-7-6, 4-3-2-A) is possible in each suit.

For that matter, mixed-suit straights are prevented as well.

White wins with the waiting move 1. Kb2! If Black moves his pawn then 2. Ng6 is mate, and if he moves his bishop then the knight captures the pawn.

Note that the king must move to b2 — moving to a2 or b1 leaves him vulnerable to a check from the bishop.

From Svenska Dagbladet, July 9, 1922.