Pandigital Primes

The digits 123456789 can be arranged to form 362,880 distinct 9-digit numbers.

How many of these are prime?

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None. The sum of the digits of each of these numbers is 45, so all are divisible by 9.

February 10, 2013February 15, 2013 | Puzzles

Black and White

brown chess problem

By Theodore Morris Brown. White to mate in two moves.

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White wins with 1. Qa8!, threatening to take the bishop with mate. There’s no escape for Black: If the bishop moves to b1 then the rook takes it, giving mate, and if 1. … Kxa1 or 1. … Bxb3 then the queen mates from h8. From The Torch, 1878.

February 8, 2013May 6, 2013 | Puzzles

Black and White

shinkman chess problem

By William Shinkman. White to mate in two moves.

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1. Rb5! and White will mate along the eighth rank. From Deutsche Schachzeitung, 1874.

February 1, 2013February 8, 2013 | Puzzles

Round Numbers

Alan and Bob are playing a game of marbles. Alan has two marbles, Bob has one, and each rolls to try to come nearest to a fixed point. If the two have equal skill, what is the chance that Alan will win?

There seem to be two contradictory arguments. On the one hand, each of the three marbles has an equal chance of winning, and two of them belong to Alan, so it seems that there’s a 2/3 chance that Alan will win.

On the other hand, there are four possible outcomes: (a) both of Alan’s rolls are better than Bob’s, (b) Alan’s first roll is better than Bob’s, but his second is worse, (c) Alan’s first roll is worse than Bob’s, but his second is better, and (d) both of Alan’s rolls are worse than Bob’s. In 3 of the 4 cases, Alan wins, so it appears that his overall chance of winning is 3/4.

Which argument is correct?

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The first argument is correct. Alan’s chance of winning is 2/3. What’s wrong with the second argument? It’s true that there are four possible outcomes, but (a) subsumes two possibilities — Alan’s first roll might come closest to the mark, with his second roll in second place, or the reverse might be true. Similarly, case (d) covers two possible outcomes — in one, Alan’s first marble takes second place and his second comes in last, and in the other these positions are reversed. So in fact there are six possible outcomes, with (a) and (d) each twice as likely as (b) or (c), and 4 of the 6 outcomes favoring Alan. From J. Bertrand, Calcul des Probabilités, 1889, via Eugene Northrop, Riddles in Mathematics, 1975.

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January 30, 2013February 8, 2013 | Puzzles

More Is Less

A riddle attributed to British prime minister George Canning, among many others:

A word there is of plural number,
Foe to ease and tranquil slumber.
Any other word you take
And add an S will plural make;
But if you add an S to this,
So strange the metamorphosis,
Plural is plural now no more
And sweet what bitter was before.

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CARES. (For those who hate trigonometry, another answer is COSINES.)

January 28, 2013February 8, 2013 | Puzzles

Black and White

taylor chess problem

By I.O. Howard Taylor. White to mate in two moves.

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1. Qh1! and White will mate with 2. Qa8 or 2. Kf7. From The Westminster Papers, 1872.

January 24, 2013February 8, 2013 | Puzzles

Half and Half

half and half puzzle

Choose an arbitrary point inside an equilateral triangle and draw segments to the vertices and perpendiculars to the sides. This divides the triangle into six smaller triangles, A, B, C, D, E, and F. Prove that the areas A + C + E and B + D + F are equal.

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Draw three additional lines through the chosen point, each parallel to a side of the large triangle. This divides it into three parallelograms and three equilateral triangles. The parallelograms are bisected by their diagonals and the triangles by their altitudes, so A + C + E = x + a + y + b + z + c = B + D + F. From Quantum, January 1990, via Ross Honsberger, Mathematical Chestnuts From Around the World, 2001.

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January 22, 2013February 8, 2013 | Puzzles

Black and White

schachmatnyi listok chess problem

From Schachmatnyi Listok, 1861. White to mate in two moves.

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1. Qh8! and the queen will mate next move.

January 17, 2013February 8, 2013 | Puzzles

Hostilities

A 10×10 chessboard contains 41 rooks. Prove that there are five rooks that don’t attack one another.

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Roll the board into a cylinder. The cylinder has 10 diagonals; because there are 41 rooks, some five of them must occupy one diagonal, and thus do not attack one another. From Alexander Soifer and Edward Lozansky, “Pigeons in Every Pigeonhole,” Quantum, January 1990.

January 16, 2013February 8, 2013 | Puzzles

Square Deal

square deal - problem

A “coffin,” or killer problem, from the oral entrance exams to the math department of Moscow State University:

Construct (with ruler and compass) a square given one point from each side.

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If A, B, C, and D denote the four points, connect A and C and draw a perpendicular from B to this segment. Let D’ lie on this perpendicular such that AC = BD’. Now D’ lies on the same side of the square (or on an extension of that side) as D. Connecting D and D’ gives us the orientation of the square, and the rest is easy. According to Tanya Khovanova and Alexey Radul, in the 1970s such problems were cultivated expressly to prevent Jews and other undesirable students from entering the university. The challenging nature of the problems prevented students from solving them, and the existence of a simple solution gave cover to the administration. For further examples, see http://arxiv.org/pdf/1110.1556v1.pdf.

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January 14, 2013February 8, 2013 | Puzzles · Science & Math