Yes. The father sets out with his first son, leaving the second boy to walk alone. Father and son drive 24 kilometers, which takes 6/5 hours. Then the father puts down the first son and rides back to meet the second son 9 kilometers from home, which takes 3/5 hours. Finally he and the second son ride 6/5 hours more and reach grandmother’s house just as the first son is arriving there on foot. Each son spends 6/5 hours riding 24 kilometers and 9/5 hours walking 9 kilometers, so the three of them reach grandmother’s in exactly 3 hours.

The surprising 1. Qh3! leaves Black with no options. If he moves his knight then White can take the rook with mate, and if he moves the rook then White’s knight will mate.

From Teplitz-Schönauer Anzeiger, 1921.

“The two children were so befogged over the calendar that they had started on their way to school on Sunday morning!”

In presenting this problem in Ingenuity in Mathematics, Ross Honsberger gave a complex analysis, but his friend Scott Vanstone showed a simple way to visualize it, building on an observation by John Annulis:

Each boy is represented by a point, and each pair of boys named in an accusation are joined by a line. Now we know that each line has a thief at one end and an innocent at the other, except for one line, which has an innocent at both ends (due to the outright liar). Since four boys named one guilty boy correctly, the two points that represent the thieves are attached to a total of four lines. And because no boy made two accurate accusations, the thieves themselves are not joined by a line. The only two points that satisfy all these requirements indicate Charlie and James.

Yes. First three cannibals cross; then the two rowers return, bearing an extra missionary; then the rowing missionary crosses with a cannibal and returns with two missionaries; and finally the rowing cannibal crosses:

Mmmm Cccc
MmmmCcc c
mmcc MmCc
Mmmccc mC
ccc MmmmC
Cccc Mmmm

1. Qf8! and the queen will mate.

From the Chicago Times, 1879.

80 cents. It’s tempting to say that the answer is 81 cents, since paying for nine 90-cent packets will give us 10 packets total, meaning we’ve paid ($0.90 × 9) / 10 = $0.81 for each packet. But the free packet itself includes a voucher, which is worth something.

Maslanka suggests this thought experiment: Borrow a packet from a friend, then buy eight additional packets for $0.90 × 8 = $7.20. Consume the nine packets, redeem the nine vouchers for a free packet, and use this to repay your friend. You’ve consumed nine packets of sugar for $7.20, so each packet is worth 80 cents.

From Maslanka’s “A Bouquet of Brainteasers,” in Tribute to a Mathemagician, 2005.

Take 1 pill from the first bottle, 2 from the second, 4 from the third, 8 from the fourth, 16 from the fifth, and 32 from the sixth. This gives us 63 pills that ought to weigh 315 grams. As before, the bad pills will add some excess weight, but now that weight can tell us which combination of bottles is bad. Suppose the 63 pills weigh 336 grams, or 21 grams too much. That extra 21 grams can only have come from the fifth bottle (16 grams), the third (4 grams), and the first (1 gram). The same is true for any overage: Because any integer is uniquely expressible as the sum of powers of 2, the excess weight will always identify uniquely any combination of bad bottles.

He numbers the bottles from 1 to 10, then takes one pill from bottle one, two pills from bottle two, and so on. This gives him a pile of 55 pills that ought to weigh 275 grams. Its difference from this weight can tell him which bottle is bad: If the pile is 1 gram overweight then the bad pills are in bottle one, if 2 grams then they’re in bottle two, and so forth.

1. Ra5! The rook tours the fifth rank and returns to its original square, forcing 1. … Kc1 2. Ra1#. The normal-seeming 1. Rxa6, which would ordinarily lead to the same mate, here fails because Black can take the rook.

By Kuznecov and Plaskin, from Miodrag Petković, Mathematics and Chess, 1997.