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From the 2001 Moscow Mathematical Olympiad:

Before you are three piles of stones. One contains 51 stones, one 49 stones, and one 5 stones. On each move you can either combine two piles into one or divide any pile with an even number of stones into two equal piles. Is it possible to end up with 105 piles, each containing a single stone?

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No. If at any point the number of stones in each pile is a multiple of the same odd number a, then this will remain true after any number of operations. On the first move we’re forced to create two piles of 100 and 5 stones (both divisible by 5), or 56 and 49 stones (both divisible by 7), or 51 and 54 stones (both divisible by 3). Whichever course we choose, all subsequent operations will produce piles divisible by that odd factor. This prevents us from producing piles of single stones.

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August 1, 2013August 9, 2013 | Puzzles

An Invitation

On Nov. 25, 1862, Abraham Lincoln sent this dispatch to Gen. Ambrose Burnside at Aquia Creek, Va.:

Can Inn Ale me withe 2 oar our Ann pas Ann me flesh ends NV Corn Inn out with U cud Inn heaven day nest Wed roe Moore Tom darkey hat Greek Why Hawk of abbott Inn B chewed I if.

What did it mean?

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David Homer Bates, manager of the War Department Telegraph Office, explains: “By reading the above backward, observing the phonetics, and bearing in mind that flesh is the equivalent of meat, the real meaning is easily found”: If I should be in a boat off Aquia Creek at dark tomorrow (Wednesday) evening, could you without inconvenience meet me and pass an hour or two with me? — A. Lincoln “It cannot be said that this specimen exhibits specially clever work on the part of the War Department staff, nor is it likely that the Confederate operator, if he overheard its transmission, had much trouble in unraveling its meaning. As to this we can only conjecture.” Burnside understood it, and the two met. (David Homer Bates, “Lincoln in the Telegraph Office,” The Century Magazine, June 1907)

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July 28, 2013August 9, 2013 | Puzzles

Dictation

In 1887, Irish journalist Richard Pigott sold a series of letters to the Times of London. Purportedly written by Irish Parliamentary Party leader Charles Parnell, they seemed to show that Parnell had approved of a savage political assassination five years earlier. Parnell denounced the letters as a “villainous and bare-faced forgery.”

In the ensuing investigation, Parnell’s attorney asked Pigott to write a series of words and submit them to the court:

livelihood
likelihood
Richard Pigott
proselytism
Patrick Egan
P. Egan
hesitency

What was the point of this?

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Pigott misspelled hesitancy … as had the forger. He fled to the continent, sent a full confession from Paris, and shot himself in a hotel room in Madrid.

July 27, 2013August 2, 2013 | History · Puzzles

Black and White

ognjanovic chess problem

By Zarko Ognjanovic, 1912. White to mate in two moves.

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1. Kc5! followed by 2. Qa8# or 2.axb6#.

July 26, 2013August 2, 2013 | Puzzles

Buoy Howdy

From Russian puzzle maven Boris Kordemsky:

Two diesel ships leave a pier at the same time. One travels upstream, the other downstream, each with the same motive power. As they depart, one drops a lifebuoy into the water. An hour later, both ships reverse course. Which will reach the buoy first?

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Consider the situation from the perspective of the floating buoy. Now there is no current; the two ships move away from the buoy at equal speeds and return at equal speeds. They’ll reach it simultaneously.

July 25, 2013August 2, 2013 | Puzzles

Square Dance

paint scheme puzzle

From the 2001 Moscow Mathematical Olympiad:

Two stones, one black and one white, are placed on a chessboard. A move consists of moving one stone up, down, left, or right. The two stones may not occupy the same square. Does a sequence of moves exist that will produce every possible arrangement of the stones, each occurring exactly once?

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No. Call an arrangement even if the stones occupy squares of the same color, odd otherwise. In the sequence we’re seeking, the number of odd arrangements can differ from the number of even arrangements by at most 1. The number of odd arrangements is 64 × 32 — choose any square for the black stone and then any square of the opposite color for the white. But the number of even arrangements is 64 × 31, since the two stones can’t occupy the same square. So the task is impossible.

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July 22, 2013August 2, 2013 | Puzzles

The Puzzle of the Self-Torturer

You’re fitted with a watch that imparts an electric current to your skin in increments too small to distinguish. Initially it’s set to 0 (off), and the settings run up to 1000. At the start of each week you’re allowed a period of experimentation to compare various settings, and then the watch is returned to its last setting. Then you have the option to increase the setting by 1; if you do this, you get $10,000. You may never reduce the setting.

What should you do? On the first day your experimentation shows you that the highest setting is completely intolerable; at that setting you’d pay any amount of money to get rid of the watch. But on this first week your decision is simply whether to advance from 0 to 1, getting $10,000 for accepting an imperceptible amount of pain. That seems attractive.

The trouble seems to be that your evaluations are “transitive” only at a large scale. If you prefer 0 to 500 and 500 to 1000, then it’s valid to conclude that you’ll prefer 0 to 1000. But if you prefer 51 to 4 (because of the financial reward) and 103 to 51, can we conclude that you’ll prefer 103 to 4? Not necessarily.

Unfortunately for all of us, this describes a lot of life. “The self-torturer is not alone in his predicament,” writes philosopher Warren S. Quinn, who proposed this puzzle in 1990. “Most of us are like him in one way or another. We like to eat but also care about our appearance. Just one more bite will give us pleasure and won’t make us look fatter; but very many bites will. And there may be similar connections between puffs of pleasant smoking and lung cancer, or between pleasurable moments of idleness and wasted lives.” What’s the best course?

(Warren S. Quinn, “The Puzzle of the Self-Torturer,” Philosophical Studies, May 1990)

July 21, 2013July 20, 2013 | Puzzles

Open and Shut

pearson shutter puzzle

A poser by Cyril Pearson, puzzle editor for the London Evening Standard at the turn of the 20th century:

Upon the shutters of a barber’s shop the legend above was painted in bold letters. One evening about 8:30, when it was blowing great guns, quite a crowd gathered round the window, and seemed to be enjoying some excellent joke. What was amusing them when the shutter blew open?

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The shutters were divided vertically just to the left of the I in the top line; cover up the left side of the legend and read the message I SHALL DANCE DAILY AND UTTER LOUD SCREAMS IN THIS WINDOW LISTEN AFTER 8 P.M.

July 21, 2013July 26, 2013 | Puzzles

Babel on Olympus

In 1996, MIT philosopher George Boolos published this puzzle by Raymond Smullyan in The Harvard Review of Philosophy, calling it “the hardest logical puzzle ever”:

Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.

Boolos made three clarifying points:

It could be that some god gets asked more than one question (and hence that some god is not asked any question at all).
What the second question is, and to which god it is put, may depend on the answer to the first question. (And of course similarly for the third question.)
Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.

What’s the solution?

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Boolos’ original article includes his own solution; the ensuing years have seen a lively search for improvements. The solution below, by Brian and Landon Rabern, simplifies matters by using embedded questions (“If I asked you x, would you say y?”): Question 1: Ask God B, “If I asked you ‘Is A Random?’, would you say ja?” — If B answers ja, then either B is Random or his answer tells us that A is Random. — If B answers da, then either B is Random or his answer tells us that A is not Random. In either case, God B’s answer to Question 1 has identified a god who is not Random. Question 2: Ask the god who is not Random, “If I asked you ‘Are you False?’, would you say ja?” — If he answers da, then he is True. — If he answers ja, then he is False. Question 3: Ask the same god, “If I asked you ‘Is B Random?’, would you say ja?” — If he answers ja, then B is Random. — If he answers da, then the third god is Random. At this point we’ve established the identities of two of the gods, so we can deduce the third by elimination.

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