If n is odd then the sequence (1, 2, 3, … n) begins and ends with an odd number. This means it contains one more odd number than even. Likewise (a₁, a₂, … a_n), which are only the same numbers in a different order. This means that two odd numbers must end up in the same bracket — there are not enough even numbers available to pair each with an odd. And if two odd numbers appear in the same bracket, they’ll produce an even difference and hence an even product.

From Ross Honsberger, Ingenuity in Mathematics, 1970.

Three. Three men can be mutual brothers-in-law if A marries B’s sister, B marries C’s sister, and C marries A’s sister. But D has no way to join this group. He cannot ask his sister to marry A, B, or C, because they are already spoken for. And if he himself marries the sister of, say, A, then he cannot be a brother-in-law to C unless C or D is A’s brother, which implies a violation of the law against consanguinity. (If a man may marry his own sister, then there’s no limit to the number of mutual brothers-in-law.)

(David L. Silverman, “A Problem of Relations,” Journal of Recreational Mathematics 3:3, July 1970)

They’ll form a circle:

1. Qd2 and Black can’t avoid a fatal discovered check.

Two widow Ladies married were,
Each to the other’s Son;
And they both pregnant did appear
E’er one full Year was run.
The Consequence of which did prove,
To each a charming Boy;
This did cement their Husband’s Love,
And added to their Joy.
By this Event likewise it’s plain
They did commence Grandmothers,
And that their Husbands did obtain
Two young delightful Brothers.
A Brother you may justly call
Each, to the other’s Father:
Uncles they were reciprocal,
You easily may gather.

No. Eliminate the even numbers and those ending in 5. The sum of the digits in each of the remaining numbers will always produce a multiple of 3, so the numbers themselves will always be divisible by 3.

The resourceful 1. Qe5 threatens 2. Qb8#, and now:

1. … Rxe5 2. b7#
1. … Rd6 2. Qe8#
1. … Re8 2. Qxe8#
1. … Rxb6+ 2. Nxb6#
1. … Nc6 2. b7#

Imagine taking a practice run in a car that has plenty of gas. Along the way the car will pick up exactly enough gas to take it once around the track, so it will complete the practice run with the same amount of fuel that it started with. If we keep an eye on the car’s fuel gauge, we’ll find that it reaches its lowest point as we arrive at one of the depots (call it x_i). Any additional gas in our tank at that point is unneeded; we know it will never be used. Now if we start at x_i with an empty tank we know we’ll have enough gas to get from each depot to the next, running out of gas just as we arrive back at x_i.

At first it sounds impossible, but the task can be completed within five moves:

1. After the first spin, reach into wells in diagonally opposite corners of the table and set both tumblers upright.
2. Let the table spin again and reach into two adjacent wells. Because of move 1, you’ll find that at least one of them is upright. If the other is inverted, turn it upright. If the bell doesn’t sound, then the position is now UUUD.
3. After the next spin, reach into diagonally opposite corners. If you find an inverted tumbler, turn it up and the bell will ring. If both tumblers are upright, invert one, leaving the position UUDD (two inverted tumblers in adjacent wells).
4. Let the table spin, then choose two adjacent wells and reverse both tumblers. If they were in the same position, then the bell will sound. Otherwise we know the position is now UDUD.
5. After the fifth spin, reach into diagonally opposite wells and reverse both tumblers, and the bell must sound.

No wise young pigs go up in balloons.

(Probably true.)