Falling Gravity

A water jug is empty, and its center of gravity is above the inside bottom of the jug. Water is poured into the jug until the center of gravity of the jug and water (considered together) is as low as possible. Explain why this center of gravity must lie at the surface of the water.

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If P is the center of gravity of the system, then as long as P is above the surface of the water, it will fall as the surface rises. When P reaches the surface, adding any more water will raise the water level and hence the level of P. From Edward Barbeau, Murray Klamkin, and William Moser, Five Hundred Mathematical Challenges, 1995.

March 8, 2015March 15, 2015 | Puzzles

Time and Distance

A puzzle from Martin Gardner’s column in Math Horizons, November 1995:

Driving along the highway, Mr. Smith notices that signs for Flatz beer appear to be spaced at regular intervals along the roadway. He counts the number of signs he passes in one minute and finds that this number multiplied by 10 gives the car’s speed in miles per hour. Assuming that the signs are equally spaced, that the car’s speed is constant, and that the timed minute began and ended with the car midway between two signs, what is the distance from one sign to the next?

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We can answer this without knowing the car’s speed. If x is the number of signs that the car passes in one minute, then the car will pass 60x signs in an hour. We’re told that the car is traveling at 10x miles per hour, so in 10x miles it will pass 60x signs, and in one mile it will pass 60x/10x signs, or 6. So the signs are 1/6 mile, or 880 feet, apart.

March 7, 2015March 15, 2015 | Puzzles

Lineup

A group of children are standing outside a room. Each wears a hat that’s either red or blue, and each child can see the other children’s hats but not her own. At a signal they enter the room one by one and arrange themselves in a line partitioned by hat color. How do they manage this without communicating?

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The first child stands in the center of the room, and the second stands at her side. Thereafter each entering child follows a rule: If she sees that the line of children are all wearing hats of the same color, then she stands at one end of the line. If they’re wearing both red and blue hats, then she inserts herself at the break between the two colors. (Thanks, Matan.)

March 5, 2015March 15, 2015 | Puzzles

Black and White

loyd chess problem

By Sam Loyd. White to mate in two moves.

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1. Qb4! and the queen will mate.

March 4, 2015March 15, 2015 | Puzzles

Book Codes

Benedict Arnold encrypted his messages to the British Army using Blackstone’s Commentaries on the Laws of England. Arnold would replace each word in his message with a triplet of numbers representing the page number, line number, and word position where the word might be found in Blackstone. For example:

The 166.8.11 of the 191.9.16 are 129.19.21 266.9.14 of the .286.8.20, and 291.8.27 to be on 163.9.4 115.8.16 114.8.25ing — 263.9.14 are 207.8.17ed 125.8.15 103.8.60 from this 294.8.50 104.9.26 — If 84.8.9ed — 294.9.12 129.8.7 only to 193.8.3 and the 64.9.5 290.9.20 245.8.3 be at an 99.8.14.

British Army Major John André could then look up the words in his own copy of Blackstone to discover Arnold’s meaning:

The mass of the People are heartily tired of the War, and wish to be on their former footing — They are promised great events from this year’s exertion — If disappointed — you have only to persevere and the contest will soon be at an end.

The danger in using a book code is that the enemy can decode the messages if he can identify the book — and sometimes even if he can’t. In the comic strip Steve Roper, a reporter once excitedly telephoned the coded message 188-1-22 71-2-13 70-2-11 68-1-25 19-1-6 112-2-10 99-1-35. Reader Sean Reddick suspected that this message had been encoded using a dictionary, with each triplet of numbers denoting page, column, and word number. He never did discover the book that had been used, but by considering the ratios involved and consulting half a dozen dictionaries he managed to break the code anyway — he sent his solution to a nationally known columnist, who verified his feat when the comic strip bore out his solution. What was the message? (Hint: In the comic, the reporter mentions significantly that the plaintext message was given to him by “the delivery boy.”)

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WOMAN HERE HAS GIVEN BIRTH TO SEXTUPLETS. This appears in Dave Silverman’s article “Some Cryptographic Challenges” in the May 2008 issue of Word Ways; I’m not sure who the columnist was. Silverman’s own best guess was WISCONSIN MUSICAL MULTIMILLIONAIRE LEADS CONCERT UNDER PAVILION, which would have made for an entertaining comic.

March 2, 2015March 8, 2015 | Puzzles

One Two Three

Each point on a straight line is either red or blue. Show that it’s always possible to find three points of the same color in which one is the midpoint of the other two.

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Choose two points of the same color, say blue. Call these A and B, and let C be their midpoint. If C is blue then we’re done. If C is red then mark points D and E as shown such that AD = AB = BE. Now if D or E is blue then we’re done, because these produce the blue triples DAB and ABE. And if D and E are both red then we have DCE, a red triple. So there’s no way to assign colors to these five points without producing some monochromatic triple.

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March 1, 2015March 8, 2015 | Puzzles

There and Back Again

John and Mary drive from Westville to Eastville. John drives the first 40 miles, and Mary drives the rest of the way. That afternoon they return by the same route, with John driving the first leg and Mary driving the last 50 miles. Who drives the farthest, and by what distance?

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We don’t know the distance between Westville and Eastville, but suppose it’s only 50 miles. In that case John drives 40 miles altogether and Mary drives 60. If the distance is any greater than that, then the two drivers are dividing the extra miles equally between them, with Mary driving east and John west. So the total distance is immaterial — Mary drives 20 more miles than John.

February 27, 2015March 8, 2015 | Puzzles

The Sixth Cent

the sixth cent puzzle

You toss 6 fair coins, and I toss 5 fair coins. What is the probability that you get more heads than I do?

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Either you’ll toss more heads than I do or you’ll toss more tails than I do, but not both. (Right?) These outcomes are symmetric, so each has probability 1/2. UPDATE: A number of readers questioned the validity of this solution, but I think that’s due to a bad writeup on my part. I don’t know who devised the problem originally; I found it in the 1995 collection Five Hundred Mathematical Challenges, by Edward Barbeau, Murray Klamkin, and William Moser. Here’s their presentation: John tosses 6 fair coins, and Mary tosses 5 fair coins. What is the probability that John gets more “heads” than Mary? More generally, suppose John tosses n + 1 coins and Mary tosses n coins. Then, either John tosses more heads than Mary does or John tosses more tails than Mary does — but not both. (Check this out!) Since these two outcomes are symmetric, each occurs with probability 1/2. I think I confused the issue by failing to note that this reasoning addresses specifically the case of n coins vs. n + 1. I still haven’t had time to think about this carefully, but most of the people who’d objected wrote back later to change their minds. If you still think it’s incorrect, please do let me know, and I’ll post further updates here. Sorry for the confusion, and thanks to everyone who’s written in. UPDATE #2: Yes, the solution appears to be valid with this change. Thanks to everyone who contributed!

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February 23, 2015August 12, 2019 | Puzzles

Black and White

brownson chess problem

By O.A. Brownson Jr. White to mate in two moves.

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1. Qh8! and the queen will mate. From Chess Problems, 1876.

February 18, 2015March 1, 2015 | Puzzles

The Three Hats Game

Three players enter a room, and a maroon or orange hat is placed on each one’s head. The color of each hat is determined by a coin toss, and the outcome of one toss has no effect on the others. Each player can see the other players’ hats but not his own.

The players can discuss strategy before the game begins, but after this they may not communicate. Each player considers the colors of the other players’ hats, and then simultaneously each player must either guess the color of his own hat or pass.

The group shares a $3 million prize if at least one player guesses correctly and no player guesses incorrectly. What strategy will raise their chance of winning above 50 percent?

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A player who sees two hats of the same color (say, orange) guesses the opposite color (maroon). A player who sees hats of two different colors passes. There are 8 possible arrangements of hats: OOO OOM OMM MOO MMO MMM MOM OMO In six of these, two of the players see hats of different colors and so will pass, and the third player sees two hats of the same color and guesses the opposite color, winning. In the other two cases, all three players are wearing hats of the same color, so all guess the wrong color and lose. This strategy produces a win in six of the eight cases, so the players will win 3/4 of the time. From “Applications of Recursive Operators to Randomness and Complexity,” the 1998 UCSB doctoral thesis of computer scientist Todd Ebert.

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February 17, 2015February 22, 2015 | Puzzles