Black and White

meredith chess problem

By William Meredith. White to mate in two moves.

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Black’s king is immobilized; his h6 knight must maintain its guard against Nxf7#; and moving his pawn would permit Rh7# (or Nxg6#). That means that only the e4 knight is free to move safely. With 1. Qb1! white immobilizes that piece as well, as moving it would now permit 2. g7#, since the queen now guards the king’s only escape square. So Black is out of safe moves, and must invite a mate. From Checkmate, 1901.

November 8, 2013November 15, 2013 | Puzzles

Flip Sum

A problem from the 1999 St. Petersburg City Mathematical Olympiad:

Fifty cards are arranged on a table so that only the uppermost side of each card is visible. Each card bears two numbers, one on each side. The numbers range from 1 to 100, and each number appears exactly once. Vasya must choose any number of cards and flip them over, and then add up the 50 numbers now on top. What’s the highest sum he can be sure to reach?

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The highest number he can be sure to reach is (1/2)(1 + 2 + 3 + … + 100), or 2525. If the visible numbers already meet this total, he should accept them as they are; otherwise he should flip all 50 cards. This might be the best that he can do. Suppose that the cards show the numbers from 26 to 75 (totaling 2525) and that the k cards he flips over are the first k cards in the sequence 1, 2, …, 25, 76, 77, …, 100. If he flips over fewer than 26 cards his sum will decrease; if he flips more than that, the sum will rise again but will reach 2525 only with the 50th card. Either way, he won’t beat that total.

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November 7, 2013November 15, 2013 | Puzzles

Black and White

larsen chess problem

By Peder Andreas Larsen. White to mate in two moves.

(I’ve added a guide to chess notation.)

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White wins with the ridiculous 1. Qg1!, threatening 2. N4b6#. If Black moves his bishop, then 2. Qa7 is mate, and if Black advances his b-pawn, then the queen takes the rook. From Skakbladet, 1918.

November 1, 2013November 8, 2013 | Puzzles

Black and White

By Francis Healey. White to mate in two moves.

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1. Qa7! and the queen or rook will mate. From A Collection of Two Hundred Chess Problems, 1866.

October 25, 2013November 1, 2013 | Puzzles

Square Deal

square deal puzzle

A puzzle by Sam Loyd. The red strips are twice as long as the yellow strips. The eight can be assembled to form two squares of different sizes. How can they be rearranged (in the plane) to form three squares of equal size?

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October 23, 2013January 30, 2022 | Puzzles

Black and White

møller chess problem

By Jørgen Thorvald Møller. White to mate in two moves.

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1. Kf6! and White will mate with 2. Qh1# or 2. Kg6#. From Skakbladet, August 1919.

October 18, 2013October 25, 2013 | Puzzles

All Relative

A problem from Dick Hess’ All-Star Mathlete Puzzles (2009):

A man points to a woman and says, “That woman’s mother-in-law and my mother-in-law are mother and daughter (in some order).” Name three ways in which the two can be related.

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The woman can be the man’s daughter-in-law, because his wife is the daughter of his mother-in-law. The woman can be the man’s wife’s aunt (his mother-in-law’s brother’s wife), because his mother-in-law is the daughter of her own brother’s wife’s mother-in-law. The woman can be the wife of the man’s nephew (the wife of his wife’s sister’s son), because his wife is the daughter of his mother-in-law. Equivalently, the man can be the woman’s father-in-law, her niece’s husband (her husband’s sister’s daughter’s husband), or her aunt’s husband (her husband’s mother’s sister’s husband).

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October 12, 2013October 18, 2013 | Puzzles

Black and White

pradignat chess problem

By Émile Leonard Pradignat. White to mate in two moves.

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1. Bd5! and White will win by Qxe6#, Qd3#, or c8=N#. From La Stratégie, 1896.

October 11, 2013October 18, 2013 | Puzzles

Ballot Measure

A Russian problem from the 1999 Mathematical Olympiad:

In an election, each voter writes the names of n candidates on his ballot. Each ballot is then placed into one of n+1 boxes. After the election, it’s noted that each box contains at least one ballot, and that if one ballot is drawn from each box, these n+1 ballots will always have a name in common. Show that for at least one box, there’s a name that appears on all of its ballots.

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Suppose, by way of contradiction, that this isn’t the case — that there is no box whose ballots all show a common name. Look at an arbitrary ballot in the first box. It lists n names; call these Candidates 1 through n. Now we can find a ballot in the second box that doesn’t list Candidate 1, a ballot in the third box that doesn’t list Candidate 2, and so on. Collect these n+1 ballots together and they form a group with no name in common. This contradicts the stated fact that no such group exists. So our assumption must be wrong: All of the ballots in at least one box must have a name in common.

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