1. Qa1 and Black must succumb:

1. … Qxa1+ 2. c3#
1. … Qa2 2. Qxc1#
1. … Qb2+ 2. Bc3#
1. … Qxb3 2. Qxc1#
1. … Qxc2 2. Qxc1#
1. … Bb2+ 2. c3#
1. … Ba3 2. c3#
1. … Bxd2 2. Qxb1#

From Good Companions, March 1920.

Fill the House and Senate arbitrarily, and then consider each member in turn. If he has two or more enemies in his current chamber, then move him to the opposite one, where he’ll have no more than one enemy. If we repeat this operation then eventually every member will have at most one enemy in his chamber.

Adapted from Svetoslav Savchev and Titu Andreescu, Mathematical Miniatures, 2003.

12/30/2011 UPDATE: Wait, in paraphrasing Savchev and Andreescu’s solution I oversimplified it. (Thanks, John.) Here’s their solution:

“Form two chambers in an arbitrary way and denote by e the total number of pairs of enemies who are members of the same chamber. If there is a parliamentarian A who has at least two enemies in his chamber, send him to the other one, where he may have no more than one enemy. So at least two pairs of enemies disappear, while no more than one new pair can be formed. It follows that e decreases at least by 1 after this operation. Repeating the same several times, we will arrive at a situation when e cannot be decreased by more (because it is a nonnegative integer). This means that each parliamentarian will have at most one enemy in his or her chamber.”

The Hatter is telling the truth. This is the only possibility that doesn’t lead to a contradiction. The Dodo lies when he says that the Hatter lies; the Hatter says correctly that the March Hare lies; and the March Hare lies when he says that both the Dodo and the Hatter lie (in fact, only the Dodo does).

1. Qa8! and the queen or the rook must mate.

From E.B. Cook et al., American Chess-Nuts, 1868.

Vowels.

B is taller:

• If A and B stand in the same row, then B is taller, since we know that B is the tallest student in his row.
• If A and B stand in the same column, then again B is taller, since we know that A is the shortest student in his column.
• If A and B share neither a row nor a column, then let C be the student who’s in the same column as A and the same row as B. Then C is shorter than B (who is the tallest in his row) and taller than A (who is the shortest in his column), so B > C > A.

In every case, then, B is taller than A.

1. Nb7, followed by 2. Qe4#.

From Le Palamède, via Charles Tomlinson, Amusements in Chess, 1845.

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)