Black and White

sam loyd chess problem

By Sam Loyd. White to mate in two moves.

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1. Be8! and Black can choose among four mates.

February 24, 2012January 31, 2013 | Puzzles

Jackson Strive

You and I spot a $20 bill on the street. To divide it, we agree to an auction: Each of us will write down a bid, and the high bidder will keep the $20 but pay the amount of his own bid to the other player. If we submit the same bid then we’ll split the $20. What should you bid?

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If you bid too low (say, $1), then in the worst case I’ll bid a penny higher (leaving you with $1.01). If you bid too high (say, $19), then in the worst case you’ll win the auction but have to give me most of your winnings. Bidding $10 guarantees you at least $10 profit: If I bid higher than I’ll have to pay you my bid, and if I bid lower then you’ll win the $20 and have to pay me only $10. But bidding $9.99 also guarantees you $10 if I bid higher — and it nets you an extra penny if I bid lower. That’s your best bet.

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February 22, 2012January 31, 2013 | Puzzles

Green Ties

The Martian Census Bureau compiled the marital history of every male and female Martian, living and dead:

Never married: 6,823,041
Married once: 7,354,016
Married twice: 1,600,897
Married three times: 171,013
Married four times: 2,682

What’s wrong with these figures?

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The total number of Martians married an odd number of times must be even (it takes two to get married, even on Mars). But 7,354,016 + 171,013 gives 7,525,029, an odd number. Contributed by Sidney Kravitz, Dover, N.J., to the Journal of Recreational Mathematics 2:4, October 1969.

February 21, 2012January 31, 2013 | Puzzles

Fields of Gold

fields of gold puzzle

Which of the yellow areas is larger?

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They’re equal. The long diagonal bisects the large rectangle as well as each of the smaller white rectangles. If the two halves of the large rectangle are equal, and if the smaller white rectangles are divided evenly between them, then the area that remains in each half (yellow) must be equal as well.

February 18, 2012January 31, 2013 | Puzzles

Black and White

From Stratagems of Chess, 1817. White to mate in two moves.

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1. Qc5+ dxc5 2. Rd8#

February 17, 2012January 31, 2013 | Puzzles

Picture This

We must assemble a jigsaw puzzle that contains 100 pieces. If a “move” consists of connecting two sections (even a section containing a single piece), what’s the minimum number of moves we must make to complete the puzzle?

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Each move reduces the total number of sections by 1. Since we start with 100 sections and end with 1, it will take 99 moves to assemble the puzzle, no matter how we proceed. See Breaking Bad.

February 14, 2012January 31, 2013 | Puzzles

Defense Posture

You walk into Arlington, Virginia. As the Pentagon comes into view, you’ll be able to see either two or three sides of the building. Which is more likely?

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They’re equally likely. Imagine someone approaching the building from the opposite side. If you can see two sides, she can see three, and vice versa. Because these are the only two possibilities, the probability of each must be 1/2. (Strictly speaking, the two probabilities are perfectly equal only if you’re approaching an abstract pentagonal building from an infinite distance. But you get the point.)

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February 11, 2012January 31, 2013 | Puzzles

Black and White

By Sam Loyd. White to mate in two moves.

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1. Rh8 Kxc2 2. Qh7#

February 10, 2012January 31, 2013 | Puzzles

Door Jam

You’re confronted with two doors. One leads to fortune, the other to death, but you don’t know which is which.

The doors are attended by two guards. One always lies, and the other always tells the truth, but again you don’t know which is which.

You can ask one question of one guard. What should you ask?

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Ask either guard, “Which door would the other guard advise me to take?” The truth-telling guard will honestly indicate the bad door, and the lying guard will falsely indicate the bad door. Either way, you’re safe in taking the opposite door. A sequel, from John Finnemore’s Souvenir Programme:

February 7, 2012January 31, 2013 | Puzzles

Spirit Reading

In a certain town, 90 percent of the residents drink coffee, 80 percent drink tea, 70 percent drink whiskey, and 60 percent drink gin. No one drinks all four. What percent of the residents drink liquor?

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Let C be the set of residents who drink coffee and C′ the set who don’t, and so on. So C′ contains 10 percent of the residents, T′ 20 percent, W′ 30 percent, and G′ 40 percent. Because every resident abstains from something, these four segments must cover 100 percent of the population. And because the four segments just total 100 percent, there can be no overlap among them. Hence every resident drinks three of the four beverages, and 100 percent drink liquor.

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February 6, 2012January 31, 2013 | Puzzles