Cheryl’s birthday is July 16.

Albert’s first statement is unsurprising — he knows only the month of Cheryl’s birthday, and each month on the list includes several possible days. But the fact that he knows that Bernard doesn’t know it is important — it tells us that the month can’t be May or June, because both of those months contain unambiguous dates. For example, if Cheryl’s birthday were May 19, then Bernard would know immediately that she was born in May, as that’s the only month whose 19th day appears on the list. Similarly, if Bernard knew the day of Cheryl’s birthday was the 18th, he’d know immediately that she was born in June. The fact that Albert can exclude these possibilities shows that he knows she wasn’t born in May or June; she must have been born in July or August.

Bernard’s statement shows that he’s followed this and now knows that the month must be either July or August. And the fact that he now knows the full birthday shows that Cheryl was born on the 15th, 16th, or 17th — if she’d been born on the 14th then he’d still be uncertain, as both July 14 and August 14 fall on the list.

Albert has narrowed the list of possibilities to July 16, August 15, and August 17. As he now declares that he knows the birthday unambiguously, it must be July 16 — if it fell in August he’d have no way to distinguish between August 15 and August 17.

No, there isn’t. Color the cells as shown:

Any path that the king takes must alternate between dark and light rooms, so the king can succeed only if there are an equal number of each (or if their numbers differ by 1). In the diagram there are 12 dark rooms and 10 light rooms, so there’s no way to visit all of them on one connected tour, unless the path travels outside the palace.

(From Tanton’s “A Dozen Questions About a Triangle,” Math Horizons 9:4 [April 2002], 23-28.)

1. Rd8! cxd5 2. Rc8#

From Brownson’s Chess Journal, May 1877.

Yes. Combine four of the half-full barrels to make two full ones. Now each son can receive three full, one half-full, and three empty barrels.

06/28/2017 A number of readers pointed out a simpler solution: Just pour half of a full barrel into an empty barrel! I should have seen this. The puzzle appears in James F. Fixx’s More Games for the Superintelligent; he doesn’t give the original source.

No. The numbers total 78, so in a successful magic square each row and column would total a third of that, or 26. But only one of the primes is even, which means that in two rows and two columns we’d have to produce an even number by adding three odd numbers, which is impossible.

If the rook weren’t on f2 then White could immediately give mate with the knight. Unfortunately, moving the rook out of the way gives Black time to play 1. … Bd4, and now White’s plan doesn’t work — if the knight checks the king then the bishop can simply capture it.

The answer is 1. Rf6! This stops the bishop from reaching d4, and Black has no way to stop 2. Nf2#.

1. Kf1! Kxh3 2. Bf5#

From La Régence, 1849.

1. Bh1! Kxh1 2. Rc1#