Countdown

Will a prime number ever appear in this series?

9
98
987
9876
98765
987654
9876543
98765432
987654321
9876543210
98765432109
987654321098
9876543210987
…

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No. Eliminate the even numbers and those ending in 5. The sum of the digits in each of the remaining numbers will always produce a multiple of 3, so the numbers themselves will always be divisible by 3.

December 6, 2011January 30, 2013 | Puzzles

Black and White

mott-smith chess problem

By Geoffrey Mott-Smith, New York Sun, 1932. White to mate in two.

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The resourceful 1. Qe5 threatens 2. Qb8#, and now: 1. … Rxe5 2. b7# 1. … Rd6 2. Qe8# 1. … Re8 2. Qxe8# 1. … Rxb6+ 2. Nxb6# 1. … Nc6 2. b7#

December 3, 2011January 30, 2013 | Puzzles

Round Trip

round trip puzzle

A problem by Hungarian mathematician Laszlo Lovász:

A track has n arbitrarily spaced fuel depots. Each depot contains a quantity of gasoline; the total amount of gas is exactly enough to take us around the track once. Prove that, no matter how the gas is distributed, there will be a depot at which an empty car can fill up, proceed around the track picking up gas at each depot, and complete a full round trip back to its starting depot.

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Imagine taking a practice run in a car that has plenty of gas. Along the way the car will pick up exactly enough gas to take it once around the track, so it will complete the practice run with the same amount of fuel that it started with. If we keep an eye on the car’s fuel gauge, we’ll find that it reaches its lowest point as we arrive at one of the depots (call it x_i). Any additional gas in our tank at that point is unneeded; we know it will never be used. Now if we start at x_i with an empty tank we know we’ll have enough gas to get from each depot to the next, running out of gas just as we arrive back at x_i.

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December 2, 2011January 29, 2013 | Puzzles

Table and Tumblers

This problem originated in Russia, according to various sources, but no one’s sure precisely where:

Before you is a square table that can rotate freely. In each corner is a deep well, at the bottom of which is a tumbler that’s either upright or inverted. You can’t see the tumblers, but you can reach into the wells to feel their positions.

Periodically the table rotates and stops at random. After each stop, you can feel two of the tumblers and turn over either, both, or neither. If all four of the tumblers are in the same state — all upright or all inverted — then a bell sounds. Otherwise the table rotates again and you make another “move.”

Can you guarantee to ring the bell in a finite number of moves? If so, how?

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At first it sounds impossible, but the task can be completed within five moves: After the first spin, reach into wells in diagonally opposite corners of the table and set both tumblers upright. Let the table spin again and reach into two adjacent wells. Because of move 1, you’ll find that at least one of them is upright. If the other is inverted, turn it upright. If the bell doesn’t sound, then the position is now UUUD. After the next spin, reach into diagonally opposite corners. If you find an inverted tumbler, turn it up and the bell will ring. If both tumblers are upright, invert one, leaving the position UUDD (two inverted tumblers in adjacent wells). Let the table spin, then choose two adjacent wells and reverse both tumblers. If they were in the same position, then the bell will sound. Otherwise we know the position is now UDUD. After the fifth spin, reach into diagonally opposite wells and reverse both tumblers, and the bell must sound.

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December 1, 2011January 29, 2013 | Puzzles

Pastorale

A logic puzzle by Lewis Carroll, July 2, 1893. What conclusion can be drawn from these premises?

All who neither dance on tight-ropes nor eat penny-buns are old.
Pigs that are liable to giddiness are treated with respect.
A wise balloonist takes an umbrella with him.
No one ought to lunch in public who looks ridiculous and eats penny-buns.
Young creatures who go up in balloons are liable to giddiness.
Fat creatures who look ridiculous may lunch in public, provided that they do not dance on tight-ropes.
No wise creatures dance on tight-ropes if liable to giddiness.
A pig looks ridiculous carrying an umbrella.
All who do not dance on tight-ropes and who are treated with respect are fat.

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No wise young pigs go up in balloons. (Probably true.)

November 29, 2011January 29, 2013 | Puzzles

Black and White

wurzburg mate in 2

Otto Wurzburg, first prize, eighth tourney, Pittsburgh Gazette-Times, June 10, 1917. White to mate in two.

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White wins with the surprising 1. Rg1!, e.g. 1. Rg1 hxg1 2. Qh5#.

November 27, 2011January 29, 2013 | Puzzles

Stormy Weather

stormy weather puzzle - 1

Stephen Barr observes that a pitched roof receives less rain per unit area than level ground does. This seems to mean that rain that falls at a slant will be less wetting than rain that falls vertically. Why isn’t this so?

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Rain falls from a cloud that’s parallel to the ground, so it’s equally wetting regardless of the angle at which it falls. The roof example doesn’t reflect this.

November 26, 2011January 29, 2013 | Puzzles

Horse Sense

horse sense puzzle

From the U.K. Schools Mathematical Challenge, a multiple-choice competition for students ages 11-14:

Humphrey the horse at full stretch is hard to match. But that is just what you have to do: move one match to make another horse just like (i.e. congruent to) Humphrey. Which match must you move?

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Move A to become the new horse’s tail:

November 19, 2011January 29, 2013 | Puzzles

Catbird Seat

catbird seat 1

A ladder is leaning against a tree. On the center rung is a pussycat. She must be a very determined pussycat, because she remains on that rung as we draw the foot of the ladder away from the tree until the ladder is lying flat on the ground. What path does the pussycat describe as she undergoes this indignity?

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Perhaps surprisingly, she describes a quarter circle whose center is at the foot of the tree. As we draw out the ladder, essentially we’re creating a series of right triangles with the same hypotenuse (the ladder). The midpoint of the hypotenuse (the pussycat) will always maintain the same distance from all three vertices, including the right angle. Another way to see this is to remove the pussycat and visualize two ladders joined by a hinge at their midpoints. These make a scissoring motion as the first ladder descends, the second ladder drawing a simple circular arc whose center is at the base of the tree.

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November 16, 2011January 29, 2013 | Puzzles

The Footsoldier

fabel footsoldier chess puzzle - solution

By Karl Fabel, from Weltspiegel, 1946. White adds a pawn and mates in two moves.

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Add the pawn at h2. Then 1. h4 gxh4 e.p. (forced) 2. Bxg6 is mate:

November 15, 2011January 29, 2013 | Puzzles