It’s a transposition cipher. Start with the letter S and skip backward three characters at a time through the string, counting spaces as characters and cycling back to the end as often as necessary:

SAY HELLO TO A GOOD BUY

In New York. Label the New York players A₁, A₂, …, A_n, and label all the others B₁, B₂, …, B_m. Because most masters live in New York, n > m. Now let the straight-line distance between two given players, A_i and B_j, be denoted A_iB_j. If we hold the tournament in New York, the total distance traveled, T, will be A₁B₁ + A₂B₂ + … + A_mB_m. But if we hold the tournament elsewhere, then the total distance traveled by each pair A_i and B_i (i = 1, 2, … m) will be at least A_iB_i. Hence the total distance traveled by all the participants will be at least T plus the sum of the distances traveled by the unpaired New York masters A_m+1, A_m+2, …, A_n.

(Proposed and solved by Chester McMaster in Pi Mu Epsilon Journal 1:8 [April 1953], 328.)

1. Nd2!

If Black moves a pawn, then White mates with the knight. If Black captures the bishop, then 2. Qd7 is mate.

Yes. Think of 10 straight lines in the plane, no two of which are parallel and no three of which meet at the same point. Let the lines be the bus routes and the points of intersection be the stops. At the start we can get from any one stop to any other, either directly (if they’re on the same line) or with just one change. If we remove one line, it’s still possible to get from any stop to any other with at most one change. But if we discard two lines, then one stop — the one at their intersection — will have no bus routes passing through it and thus become inaccessible to all the other stops.

(From A.M. Yaglom and I.M. Yaglom, Challenging Mathematical Problems With Elementary Solutions, 1964.)

03/04/2016 UPDATE: If multiple transfers are permitted, a simpler solution is to make a ring of the stops, with each successive pair connected by a route. Remove any one route and the graph remains connected, but remove any two it disconnects:

I think this may be a matter of translation. The Yagloms say that one makes a bus journey “possibly changing along the way.” Their own solution requires only one transfer.

Thanks to everyone who wrote in about this, and thanks to Herschel for the diagram.

There’s only one line that works:

1. c4+ Rxc4 2. e4+ Rxe4 3. Ne7+ Rxe7 4. Nc7+ Rxc7 stalemate.

Thirteen. Consider the next-to-last throw: Its total must be 12, 11, 10, 9, 8, or 7. If it’s 12, then the final total will be 13, 14, 15, 16, 17, or 18, all equally likely; if the next-to-last total is 11, then the final total will be 12, 13, 14, 15, 16, or 17; and so on. The only candidate to appear in every case is 13, so it’s the most likely total.

Generally, if we replace 12 by N, then the answer is N + 1.

(C.C. Carter and N.J. Fine, “The Most Likely Total,” American Mathematical Monthly 55:2 [February 1948], 98.)

Think of the stones as tennis players in a singles tournament. The heavier the stone, the better the player. Each weighing in the balance represents a game between two players, with the better (heavier) player going on to the next round of the tournament.

Since there are 32 players, the tournament will have 5 rounds, with 16, 8, 4, 2, and 1 matches. Each match produces a loser, so this system will eliminate 31 of the 32 players. The best player is unbeatable, so he’ll emerge as the winner no matter how the early rounds are arranged.

How can we find the second-best player? He didn’t win the tournament, so he must have been eliminated at some point. But the only player good enough to beat him is the strongest player, so the two of them must have faced one another in some match. If we refer to the record of the pairings, we can find the five players that the champion eliminated on his way to the top. One of these is the second-best player.

Now if these five players hold a mini-tournament among themselves, the second-best player will win it. This can be done in another four matches, for a total of 35 matches.

From the Leningrad Mathematical Olympiad, via Ross Honsberger’s Mathematical Delights, 2004.

1. Nd5! (threatening 2. Qe4#) Kf5 2.Qh5#

From Good Companion, April 1919.

As usual in chess, let the bottom left-hand square be black. If we number the rows from the bottom and the columns from the left, then every black square is either in an odd-numbered row or an even-numbered column. That is, all the black squares can be found in rows 1, 3, 5, 7 and columns 2, 4, 6, 8.

Let the black squares in rows 1, 3, 5, 7 be marked A and the black squares in columns 2, 4, 6, 8 be marked B, and let the white squares in rows 1, 3, 5, 7 be marked C. Leave the rest of the white squares unmarked. Finally, let a, b, c denote the total number of pieces that sit on squares marked A, B, C, respectively. We want to show that a + b is an even number.

Each row contains an odd number of pieces. That means that a + c, the total number of pieces in rows 1, 3, 5, 7, is the sum of four odd numbers, which means it’s even. Likewise, b + c, the total number of pieces in columns 2, 4, 6, 8, is also the sum of four odd numbers and thus even. So

(a + c) + (b + c) = a + b + 2c

is even, and this implies that a + b is even.

(From V. Proizvolov, “Problems Teaching Us How to Think,” Quantum, January-February 2002, via Ross Honsberger’s Mathematical Delights, 2004.)