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.)

1. Qb7! and the queen or knight will mate.

From Deutsches Wochenschach, 1893.

TTT = T × 111 = T × 3 × 37, so either YE or ME is 37. Either way, E is 7.

T is a digit, and T × 3 is a two-digit number ending in 7 (it’s either YE or ME, whichever of these is not 37).

So T must be 9, which means that TTT = 999 = 27 × 37, and E + M + T + Y = 2 + 3 + 7 + 9 = 21.

UPDATE: Strictly speaking we can’t assume that one of the factors must be 37 — it could be 74, a two-digit multiple of 37. But then when we consider the other factor, the only candidate that yields a product with the pattern TTT is 12, which doesn’t fit. Still, it’s necessary to check this line to be sure the solution is unique. (Thanks, Jeff and Steven.)

Of the nine nonzero digits, three require three characters (1, 2, 6), three require four characters (4, 5, 9), and three require five characters (3, 7, 8).

This means that the last digit of the number we’re looking for must be zero, because any other digit would have counterparts of equal length. EIGHTY-ONE, for instance, is the same length as EIGHTY-TWO and EIGHTY-SIX.

(It’s true that there are some unusually named numbers between 10 and 20, but it turns out that only one of these has a unique number of characters — 17, with nine — and 42 also uses nine characters.)

The tens digit must also be zero, for similar reasons: 20, 30, 80, and 90 all use six characters; and 40, 50, and 60 use five. 10 uses three characters, but this matches 1, 2, and 6. 70 uses 7 characters, but so does 15.

The discussion above regarding the units digit applies also to the hundreds digit, so that too is zero, and the candidates we’re left with are 1000, 2000, 3000, 4000, and 5000. Of these, only 3000 has a unique length, with 14 characters.

(Proposed in Pi Mu Epsilon Journal Spring 1985 by Harry Herein; solved in PMEJ Spring 1986 by Bob LaBarre.)