By Orest Jewetzky. White to mate in two moves.

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1. Nd2!
If Black moves a pawn, then White mates with the knight. If Black captures the bishop, then 2. Qd7 is mate.

March 9, 2016  Puzzles
A city has 10 bus routes. Is it possible to arrange the routes and bus stops so that if one route is closed it’s still possible to get from any one stop to any other (possibly changing buses along the way), but if any two routes are closed, there will be at least two stops that become inaccessible to one another?

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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.

March 3, 2016  Puzzles
A backward chess puzzle by Karl Fabel. What moves must White play to avoid winning?

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There’s only one line that works:
1. c4+ Rxc4 2. e4+ Rxe4 3. Ne7+ Rxe7 4. Nc7+ Rxc7 stalemate.

February 23, 2016  Puzzles
A fair die bearing the numbers 1, 2, 3, 4, 5, 6 is repeatedly thrown until the running total first exceeds 12. What’s the most likely total that will be obtained?

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Thirteen. Consider the nexttolast 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 nexttolast 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.)

February 18, 2016  Puzzles
If we’re given 32 stones, each a different weight, how can we find both the heaviest and the second heaviest stone in 35 weighings with an equalarm balance?

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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 secondbest 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 secondbest player.
Now if these five players hold a minitournament among themselves, the secondbest 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.

February 13, 2016  Puzzles
Henri Gerard Marie Weenink. White to mate in two moves.
February 11, 2016  Puzzles
In a certain chess position, each row and each column contains an odd number of pieces. Prove that the total number of pieces on black squares is an even number.

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As usual in chess, let the bottom lefthand 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 oddnumbered row or an evennumbered 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, JanuaryFebruary 2002, via Ross Honsberger’s Mathematical Delights, 2004.)

February 5, 2016  Puzzles
W. Langstaff offered this conundrum in Chess Amateur in 1922. White is to mate in two moves. He tries playing 1. Ke6, intending 2. Rd8#, but Black castles and no mate is possible. But by castling Black shows that his last move must have been g7g5. Knowing this, White chooses 1. hxg6 e.p. rather than 1. Ke6. Now if Black castles he can play 2. h7#.
“Not so fast!” Black protests. “My last move was Rh7h8, not g7g5, so you can’t capture en passant.”
“Very well,” says White. “If you can’t castle, then I play 1. Ke6.” And we’re back where we started.
“What was really Black’s last move?” asks Burt Hochberg in Chess Braintwisters (1999). “If a position has a history, it can have only a single history, and Black would not be able to choose what his last move was any more than I can choose today what I had for dinner last night.”
“This is not a real game, however, but a problem in chess logic. The position’s history does not exist in actuality but only as a logical construct.”
January 28, 2016  Puzzles
By Richard Steinweg. White to mate in two moves.
January 16, 2016  Puzzles
A problem from the 1973 American High School Mathematics Examination:
In this equation, each of the letters represents uniquely a different digit in base 10:
YE × ME = TTT.
What is E + M + T + Y?

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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 twodigit 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 twodigit 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.)

January 13, 2016  Puzzles