Black and White

van beek chess problem

By H. Van Beek, Chemnitzer Wochenschach, 1926 (special prize). White to mate in two moves.

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1. Nc4 and White will mate by 2. d5, 2. Qg8, or 2. Qc8.

July 20, 2012February 5, 2013 | Puzzles

Outpourings

gedanken tank puzzle 1

A tank of water has two holes of equal area, one at top and one at bottom. The top one leads to a downspout, so that both holes discharge their water at the same level. Ignoring friction, which hole produces the faster flow of water?

You actually don’t need to know the physics in order to solve this — it yields to an insight.

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The water flows from both openings at the same rate. If it didn’t, we could connect the two to create a perpetual-motion machine. (From Lewis Epstein’s Thinking Physics Is Gedanken Physics, 1989.)

July 19, 2012February 5, 2013 | Puzzles · Science & Math

Substantial

Foaled in 1773, this thoroughbred racehorse bore the unlikely name of Potoooooooo.

How was it pronounced?

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“Potatoes.”

July 16, 2012February 5, 2013 | Puzzles

Black and White

By P. Klett. White to mate in two moves.

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1. Qh6 and the queen or knight will mate.

July 13, 2012February 5, 2013 | Puzzles

Tangled Tale

shoelace puzzle

A shoelace is lying on the floor, and I’m too nearsighted to see how the lace crosses itself at points A, B, and C. If I pull on the ends, what’s the probability that it will produce a knot?

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One in four. There are three intersections, so there are eight possible ways the lace can cross itself. Of these, only two (over/under/over and under/over/under) will produce a knot.

July 10, 2012February 5, 2013 | Puzzles

Black and White

schlechter chess problem

In 1908, two years before drawing against Emanuel Lasker for the world chess championship, Carl Schlechter published this problem in the Allgemeine Sportzeitung. White to mate in two moves:

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1. Qa6 and White will mate by 2. Qh6, 2. Nd3, or 2. Qa1.

July 6, 2012February 5, 2013 | Puzzles

Togetherness

distance sums problem

What location on this line segment has the smallest sum of distances to the labeled points?

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Consider first the endpoints, A and K: The sum of the distances to these two points will be the same for any point between them. Similarly, any point between B and J will have the same total distance to points A, K, B, and J. Proceeding in this way and taking the points in pairs, we find that every point between E and G produces the same total distance to the outermost 10 labeled points. That leaves only F. So the answer is that point F has the minimum total distance to the labeled points. Strangely, this means that we can shift the points around without affecting this result. If points A through E crowded F closely on the left, and point K were 90 light-years away on the right, point F would still yield the smallest total sum, so long as the points appeared in the order shown above.

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June 30, 2012February 5, 2013 | Puzzles

Black and White

By Herbert I. Ladd. In A Selection of Popular Two-Move Chess Problems (1887), Orestes Augustus Brownson calls this “a gem of purest ray serene.” White to mate in two moves.

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1. Ng6 and Black must lose by 2. Qd3#, 2. Qe5#, or 2. Qd6#.

June 29, 2012February 5, 2013 | Puzzles

Club Med

My doctor wants to establish a dosage for a new drug, so he gives me a bottle of 48 pills and tells me to take them throughout the month of June. I can take as many or as few as I like on any given day, so long as I take at least 1 pill each day. Show that there’s a sequence of consecutive days during which I take exactly 11 pills.

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We can solve this problem by an application of the pigeonhole principle. Get a calendar for the month of June and on each day record the running total number of pills I’ve taken so far in the month. So June 1 might read “1,” June 2 “5,” and so on. The entry for June 30 will read “48.” Write these numbers in red. Now revisit each date, add 11 to the red number, and enter the result in blue. In the example above June 1 now reads “1 12,” June 2 “5 16,” and so on. June 30 reads “48 59.” Now we’ve filled the calendar with 60 numbers, all of which fall in the range 1-59. This means that at least one number falls on two different days. In the interval between these two days I took 11 pills. By similar arguments it can be shown that intervals must exist during which I took precisely 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 pills. Adapted from Kenneth R. Rebman, “The Pigeonhole Principle (What It Is, How It Works, and How It Applies to Map Coloring),” The Two-Year College Mathematics Journal, January 1979.

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June 28, 2012February 5, 2013 | Puzzles

Readership

Assume that these statements are true:

Not all Futility Closet readers are U.S. residents.
All walrus veterinarians who are not U.S. residents are not Futility Closet readers.

Do 1 and 2 imply that

Not all Futility Closet readers are walrus veterinarians?

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Yes. Statement 1 says that there are Futility Closet readers who are not U.S. residents. Let A be one of them. We’re claiming that A is not a walrus veterinarian, which would imply that statement 3 is true. Suppose the contrary, that A is a walrus veterinarian. Since he’s not a U.S. resident, statement 2 implies that he’s not a Futility Closet reader, which isn’t true. Adapted from Svetoslav Savchev, Mathematical Miniatures, 2003.

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June 27, 2012February 5, 2013 | Puzzles