Black and White

erdey chess problem

By Aladár Erdey. White to mate in two moves.

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1. Qf8! and 2. b5 or 2. Qe8 will mate. From Magyar Sakkvilág, 1911.

April 18, 2018 | Puzzles

Black and White

ferber chess problem

By Eugène Ferber, 1899. White to mate in two moves.

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1. Qc6! threatens 2. Qh1#. If the bishop moves to d3 or e4 to prevent this, then 2. Qc1 is mate.

March 29, 2018 | Puzzles

Black and White

bilfinger chess problem

By Carl Bilfinger. White to mate in two moves.

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1. Rd3! The king is forced to take the rook, and 2. Qf3 is mate.

March 6, 2018 | Puzzles

The Three-Hat Problem

In the October 2003 issue of MIT Technology Review, Donald Aucamp offered this conundrum:

Three logicians, A, B, and C, are wearing hats. Each hat displays a positive integer, and each logician can see his companions’ numbers but not his own. All of them know that the numbers are positive integers and that one of the numbers is the sum of the other two. The three then take turns in a contest to see who can determine his number first. In the first round, all three pass, but in the second round A correctly states his number is 50. What are the other two numbers, and how did A know that his was 50?

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This solution is by Howard Haber, who writes, “The key observation is that if two of the three integers are equal, then the person who sees the equal integers knows that his number is twice the same integer.” Disregard the actual values for the moment and focus on their ratio instead. For A, B, C respectively, this ratio must be 5, 2, 3. To see why, consider the first round. A, who sees the numbers 2, 3, concludes that he himself must be wearing either 1 (1 + 2 = 3) or 5 (2 + 3 = 5). A can’t get any further than this in the first round and watches the play continue, adopting for the moment the hypothesis that A, B, C are 1, 2, 3. He’s not surprised when B passes his turn, then, because that’s consistent with his idea: B would have seen A wearing 1 and C wearing 3 and been unable to decide whether he himself were wearing 2 (1 + 2 = 3) or 4 (1 + 3 = 4). So he’d pass. So far, so good. Now play passes to C. If A, B, C were 1, 2, 3, then C would see A wearing 1 and B wearing 2. That would tell C that he himself must be wearing 3 (1 + 2 = 3) or 1 (1 + 1 = 2). But unlike the first two players, C would have a way to eliminate one of his candidates: He would know that he himself could not be wearing 1, because if he were, then B would have seen the numbers 1 and 1 on his turn, and thus would have realized that his own (B’s) number must have been 2 (according to the key observation above). With that insight B would have had enough information to win the game on his turn — he could have declared that A, B, C were 1, 2, 1. But B passed his turn, revealing to C that B was unable to reach that conclusion. And seeing that, C could then have concluded that the numbers could not have been 1, 2, 1 and so must have been 1, 2, 3. But C too passes his turn. That tells A (finally!) that his starting hypothesis was wrong: A, B, C can’t be 1, 2, 3. That leaves only the second possibility he’d identified, where A, B, C are 5, 2, 3, and he announces this. “For the problem as stated [i.e., using actual values rather than ratios], just multiply all numbers above by 10.”

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This solution is by Howard Haber, who writes, “The key observation is that if two of the three integers are equal, then the person who sees the equal integers knows that his number is twice the same integer.”

Disregard the actual values for the moment and focus on their ratio instead. For A, B, C respectively, this ratio must be 5, 2, 3. To see why, consider the first round. A, who sees the numbers 2, 3, concludes that he himself must be wearing either 1 (1 + 2 = 3) or 5 (2 + 3 = 5). A can’t get any further than this in the first round and watches the play continue, adopting for the moment the hypothesis that A, B, C are 1, 2, 3. He’s not surprised when B passes his turn, then, because that’s consistent with his idea: B would have seen A wearing 1 and C wearing 3 and been unable to decide whether he himself were wearing 2 (1 + 2 = 3) or 4 (1 + 3 = 4). So he’d pass. So far, so good. Now play passes to C. If A, B, C were 1, 2, 3, then C would see A wearing 1 and B wearing 2. That would tell C that he himself must be wearing 3 (1 + 2 = 3) or 1 (1 + 1 = 2). But unlike the first two players, C would have a way to eliminate one of his candidates: He would know that he himself could not be wearing 1, because if he were, then B would have seen the numbers 1 and 1 on his turn, and thus would have realized that his own (B’s) number must have been 2 (according to the key observation above). With that insight B would have had enough information to win the game on his turn — he could have declared that A, B, C were 1, 2, 1. But B passed his turn, revealing to C that B was unable to reach that conclusion. And seeing that, C could then have concluded that the numbers could not have been 1, 2, 1 and so must have been 1, 2, 3.

But C too passes his turn. That tells A (finally!) that his starting hypothesis was wrong: A, B, C can’t be 1, 2, 3. That leaves only the second possibility he’d identified, where A, B, C are 5, 2, 3, and he announces this.

“For the problem as stated [i.e., using actual values rather than ratios], just multiply all numbers above by 10.”

March 3, 2018March 3, 2018 | Puzzles

Black and White

thompson chess problem

By Theophilus Thompson. White to mate in two moves.

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1. Qh7! and the queen will mate. From Chess Problems, 1873.

February 14, 2018February 13, 2018 | Puzzles

A Clueless Crossword

From Lee Sallows:

sallows clueless crossword

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(Thanks, Lee!)

February 13, 2018 | Puzzles

Sold!

https://www.flickr.com/photos/carleasingmadesimpletm/10723206113 — Image: Flickr

An old “puzzler” from NPR’s Car Talk:

Thirty buyers attended an auction of dozens of cars. Ten of the buyers bought fewer than 6 cars; eight bought more than 7 cars; five bought more than 8 cars; and one bought more than 9 cars.

Of the 30 buyers, how many bought 6, 7, 8, or 9 cars?

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Ten of the buyers bought fewer than 6 cars, and 1 bought more than 9. Subtract those 11 from the 30 and 19 are left. (From the Nov. 10, 2008, show.)

February 7, 2018 | Puzzles

Running Cargo

This passage is from Rudyard Kipling’s 1910 story “Brother Square-Toes.” What’s notable about the bolded section?

‘I’ll have to bide ashore and grow cabbages for a while, after I’ve run this cargo; but I do wish’ — Dad says, going over the lugger’s side with our New Year presents under his arm and young L’Estrange holding up the lantern — ‘I just do wish that those folk which made war so easy had to run one cargo a month all this winter. It ‘ud show ’em what honest work means.’

‘Well, I’ve warned ye,’ says Uncle Aurette. ‘I’ll be slipping off now before your Revenue cutter comes. Give my love to sister and take care o’ the kegs. It’s thicking to southward.’

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It contains no repeated words. At 66 words, it’s the longest such passage discovered by mathematician Mike Keith in a 2002 search of Project Gutenberg. Remarkably, the same story is notable for a second passage — 15 three-letter words in a row: “‘What?’ says Toby, ‘I thought it was Gert Schwankfelder.’ He put down his fiddle and took a good look at me. ‘Himmel!’ he says. ‘I have hit the wrong boy. It is not the new boy. Why are you not the new boy? Why are you not Gert Schwankfelder?'” (Mike Keith, “Special Word-Unit Windows in Running Text,” Word Ways 35:4 [November 2002], 262-266.)

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February 1, 2018January 31, 2018 | Puzzles

Black and White

chatillon chess problem

By Octave H. Chatillon. White to mate in two moves.

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1. Qa7! followed by 2. Qe7#. From La Presse, 1895.

January 24, 2018January 23, 2018 | Puzzles

Hope and Change

Just stumbled across this in an 1889 newspaper:

To those who love mathematics, here is a simple problem for you to figure out: A man purchased groceries to the amount of 34 cents. When he came to pay for the goods he found that he had only a $1 bill, a 3-cent piece and a 2-cent piece. The grocer, on his side, had only a 50-cent piece and a quarter. They appealed to a bystander for change, but he, although willing to oblige them, had only two dimes, a 5-cent piece, a 2-cent piece and a 1-cent piece. After some perplexity, however, change was made to the satisfaction of everyone concerned. What was the simplest way of accomplishing this?

($1 is worth 100 cents, a quarter 25 cents, and a dime 10 cents.)

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No solution is given, but this seems to work: The customer gives the $1 bill and a 2-cent piece to the grocer and the 3-cent piece to the bystander. The bystander gives the 2 dimes and the 1-cent piece to the customer and the 5-cent and a 2-cent piece to the grocer. The grocer gives the 50-cent piece to the customer and the quarter to the bystander. This leaves the customer 34 cents poorer, with 71 cents; the grocer 34 cents richer, with 109 cents; and the bystander unchanged, with 28 cents.

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January 8, 2018January 8, 2018 | Puzzles