Unless the uptown and downtown trains arrive at the station at equal intervals, the man will find himself favoring one or the other. For example, if the the uptown train arrives at 1:00, 1:10, 1:20, etc., and the downtown train arrives at 1:01, 1:11, 1:21, etc., then he’ll have a 9 times greater chance of taking an uptown train.

See The Elevator Paradox.

Consider the two people separated by the shortest distance. At the signal, these two will shoot one another. Now:

• If another player has also shot at one of these two, then we’ve established that one player was shot twice, and hence that some other player (in the whole group) lacked an antagonist and stayed dry.
• If no other player shot at these two, then we can remove them from consideration and apply the same reasoning to the remaining group: The two closest neighbors fired on one another; if another player fired on them then we have our proof; if not then remove the pair and continue.

Because we’re removing players in pairs, the group under consideration always contains an odd number of players. But it can’t drop all the way to a single player, as he would be firing on no one, and we’ve said that every player fires on someone. Hence at some point we must discover a player who is fired on twice, and thus prove that some other player was not fired on at all.

From the 1987 Canadian Mathematical Olympiad, based on a problem from the 6th All Soviet Union Mathematical Competition in Voronezh, 1966.

(a) implies that both (b) and (f) are right, which is a contradiction, so (a) can’t be right. And (c) claims that (a) is right, so (c) can’t be right either.

We’ve eliminated (a) and (c). That means that if (b) is a correct statement then (d) is accurate. But (b) contradicts (d), so (b) must be false. And since (a), (b), and (c) are all false, then (d) is false as well.

Now we’re down to (e) and (f). (f) can’t be true, because it implies simultaneously that statements (a) through (d) are false and that (e), which agrees with this, is also false, another contradiction.

That leaves (e), whose truth is confirmed by the falsity of statements (a) through (d).

1. Rd8+ and the queen will mate.

Eight. Every bishop move involves a change of rank, so painting each rank a different color will do the trick. Using fewer than eight colors won’t do, as each long diagonal covers eight ranks.

1. After the ridiculous-looking Qc6 Black must succumb to a check down the b-file or a mate at close quarters by the queen.

If -679- is divisible by 72, then it’s also divisible by 8 and by 9.

If -679- is divisible by 8, then 79- is also divisible by 8 (we can ignore the two leftmost digits because any multiple of 1000 is divisible by 8). Of all the possibilities for 79-, only 792 is evenly divisible by 8, so the last digit must be 2.

If -6792 is divisible by 9, then the sum of its digits must be a multiple of 9. That means the first digit must be 3.

If the total invoice was for $367.92, then one turkey cost $5.11.

(Via George Polya and Jeremy Kilpatrick, The Stanford Mathematics Problem Book, 2013. Thanks, Sid.)

The car runs on four tires, so it accumulated 80,000 tire-miles in total. If the five tires were worn equally, then each was used for 16,000 miles.

1. Qb4! Black must capture a knight, and the queen mates from e1 or e7.

Consider some representative cells. The warden stops at cell 7 twice, on the first and seventh trips. The divisors of 7 are 1 and 7. He stops at cell 10 four times, on the first, second, fifth, and tenth trips. The divisors of 10 are 1, 2, 5, and 10. Each door receives a quantity of visits equal to its number’s divisors.

All integers except perfect squares have an even number of divisors. So after 100 trips, only the doors whose numbers are perfect squares will remain open: cells 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.