Black and White

By W. Atkinson. White to mate in two moves.

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1. Qe4 and White wins by 2. Qxe5#, 2. c5#, 2. Qg6#, or 2. Qxd5#. From Charles F. Stubbs, Canadian Chess Problems, 1890.

August 18, 2010January 25, 2013 | Puzzles

The Cigar Puzzle

From Henry Dudeney:

Two men are seated at a square-topped table. One places an ordinary cigar (flat at one end, pointed at the other) on the table, then the other does the same, and so on alternately, a condition being that no cigar shall touch another. Which player should succeed in placing the last cigar, assuming that they each will play in the best possible manner? The size of the table top and the size of the cigar are not given, but in order to exclude the ridiculous answer that the table might be so diminutive as only to take one cigar, we will say that the table must not be less than 2 feet square and the cigar not more than 4-1/2 inches long. With those restrictions you may take any dimensions you like. Of course we assume that all the cigars are exactly alike in every respect. Should the first player, or the second player, win?

Geoffrey Mott-Smith writes, “I cannot resist narrating that I first became acquainted with this gem while reading in bed, and that like an illustrious precursor I startled the household by jumping out of bed, dancing about crying ‘Eureka! Eureka!'”

What had he seen?

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The first player must place his first cigar on end in the exact centre of the table, as indicated by the little circle. Now, whatever the second player may do throughout, the first player must always repeat it in an exactly diametrically opposite position. Thus, if the second player places a cigar at A, I put one at AA; he places one at B, I put one at BB; he places one at C, I put one at CC; he places one at D, I put one at DD; he places one at E, I put one at EE; and so on until no more cigars can be placed without touching. “I smoke ten to fifteen cigars a day,” said George Burns. “At my age I have to hold on to something.”

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August 13, 2010January 25, 2013 | Puzzles

The Potato Paradox

You have 100 pounds of Martian potatoes, which are 99 percent water by weight. You let them dehydrate until they’re 98 percent water. How much do they weigh now?

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Perhaps surprisingly, the potatoes now weigh 50 pounds. The sack of potatoes originally weighed 100 pounds, and 1 percent of its weight, or 1 pound, was non-water. If that 1 pound now makes up 2 percent of the total weight, then the total weight must be 50 pounds. Not a paradox, exactly, but many people find it counterintuitive.

August 10, 2010May 9, 2013 | Puzzles

A Chess Nut

By Sam Loyd. White to mate in two moves.

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1. Qa8! and Black must succumb to a check from a8, h8, or h6.

August 3, 2010March 1, 2013 | Puzzles

The Handicap

Zachary challenges his brother Alexander to a 100-meter race. Alexander crosses the finish line when Zachary has covered only 97 meters.

The two agree to a second race, and this time Alexander starts 3 meters behind the starting line.

If both brothers run at the same speed as in the first race, who will win?

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Alexander will win again. We know that Alexander covers 100 meters in the time it takes Zachary to cover 97. If Alexander starts 3 meters behind the starting line, then the two brothers will be neck and neck at the 97-meter mark, and Alexander will pull ahead.

August 2, 2010January 24, 2013 | Puzzles

Quitting Time

career rut alphametic

A depressing alphametic by Joseph Madachy. Each letter stands for a digit. What are the digits?

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July 28, 2010January 24, 2013 | Puzzles

Chess and Dominoes

We learned in this problem that (spoiler!) if two squares of the same color are cut out of a chessboard, the remaining 62 squares cannot be tiled by 31 dominoes.

What if the squares removed are of different colors? Is the task possible then?

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Yes, it is always possible. Impose a path on the board like this: The path covers the whole chessboard, and it alternates strictly between dark and light squares. This guarantees that if two squares of opposite colors are removed, there will always be an even number of squares in the segments between them. Thus the remaining 62 squares can always be tiled by dominoes. Proof by Ralph Gomory of IBM, cited in Ross Honsberger, Mathematical Gems, 1973.

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July 26, 2010January 24, 2013 | Puzzles

Economical

economical chess puzzle

An anonymous puzzle from the British Chess Magazine, 1993. White to mate in half a move.

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Black’s last move can only have been b7-b5. So White plays the first half of an en passant capture; that is, he moves his c5 pawn to b6 but doesn’t remove the captured pawn: That’s mate.

July 15, 2010January 24, 2013 | Puzzles

“What Does The Mirror Say?”

The above line of figures does not appear very interesting at first sight, but if one asks some charming member of the fair sex to turn it upside down and hold it to a mirror to read it, a hidden meaning becomes apparent.

— Strand, December 1908

July 14, 2010August 26, 2010 | Puzzles

Side Business

Here’s proof that one leg of a triangle always equals the sum of the other two.

ABC is our triangle. Extend it make a parallelogram, as shown, and divide the parallelogram into a grid. Obviously,

AB + BC = (AG + HJ + KL + MN) + (GH + JK + LM + NC).

Now let the grid grow increasingly fine: Instead of dividing the parallelogram into a 4×4 grid, make it 5×5, then 6×6, and so on. With each iteration, the stairstep figure described above will approximate AC more closely, and yet its total length will always equal AB + BC. Thus, at the limit, AB + BC = AC. Where is the error?

(From Henry Dudeney’s Canterbury Puzzles, via W.W. Rouse Ball’s Mathematical Recreations and Essays, 1892.)

July 3, 2010August 26, 2010 | Puzzles