Rolling

I drove this car 20,000 miles and used five tires equally in accumulating the mileage. How many miles’ wear did each tire sustain?

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The car runs on four tires, so it accumulated 80,000 tire-miles in total. If the five tires were worn equally, then each was used for 16,000 miles.

March 12, 2012January 31, 2013 | Puzzles

Black and White

By J.H. Graham. White to mate in two moves.

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1. Qb4! Black must capture a knight, and the queen mates from e1 or e7.

March 9, 2012January 31, 2013 | Puzzles

Open and Shut

A warden oversees an empty prison with 100 cells, all closed. Bored one day, he walks through the prison and opens every cell. Then he walks through it again and closes the even-numbered cells. On the third trip he stops at every third cell and closes the door if it’s open or opens it if it’s closed. And so on: On the nth trip he stops at every nth cell, closing an open door or opening a closed one. At the end of the 100th trip, which doors are open?

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Consider some representative cells. The warden stops at cell 7 twice, on the first and seventh trips. The divisors of 7 are 1 and 7. He stops at cell 10 four times, on the first, second, fifth, and tenth trips. The divisors of 10 are 1, 2, 5, and 10. Each door receives a quantity of visits equal to its number’s divisors. All integers except perfect squares have an even number of divisors. So after 100 trips, only the doors whose numbers are perfect squares will remain open: cells 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.

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March 6, 2012January 31, 2013 | Puzzles

Saw 6

We want to cut a 3-inch cube into 27 1-inch cubes. We can do this by making six cuts, as shown. Can we accomplish the task with fewer cuts by rearranging the pieces between cuts?

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No. Imagine the 1-inch block at the heart of the 3-inch block. It has six faces, and all six must be freed. This requires six cuts.

March 4, 2012January 31, 2013 | Puzzles

Black and White

By T. Herlin. White to mate in two moves.

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1. Qh1 and a rook must abandon the king’s protection.

March 2, 2012January 31, 2013 | Puzzles

All Aboard

One hundred people board a 100-seat airplane. The first one has lost his boarding pass, so he sits in a random seat. Each subsequent passenger sits in his own seat if it’s available or takes a random unoccupied seat if it’s not.

What’s the probability that the 100th passenger finds his seat occupied?

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It’s 50 percent. When the 100th passenger boards, the last available seat must be either the one assigned to the first passenger or the one assigned to the last. Nothing favors either of these outcomes; they’re equally likely.

February 29, 2012January 31, 2013 | Puzzles

Making Ends Meet

making ends meet puzzle

Fifty coins of various denominations are arranged in a row. You will take a coin from either of the row’s ends, then I will, and so on until all the coins are gone. What strategy will ensure that you take at least as much money as I?

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Before you begin, number the coins from 1 to 50 and determine whether the even-numbered or odd-numbered coins yield a higher sum. Because you’re moving first, you can contrive to capture all of those coins. For example, if the even-numbered coins are more valuable, begin by taking coin 50. That forces me to take 1 or 49, which permits you to take 2 or 48, and so on. From Peter Winkler’s excellent book Mathematical Puzzles: A Connoisseur’s Collection, 2004.

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February 27, 2012January 31, 2013 | Puzzles

Half and Half

half and half puzzle

It’s easy to draw a line and scatter an equal number of points on either side. Is it always possible to do this in reverse order? That is, given a finite set of points in the plane, is it always possible to draw a line that divides it neatly in two? (If there are an odd number of points, assume the line must intercept precisely one of them.)

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Yes, it’s always possible. Join every pair of points with a line segment. Now, to the left of the field, draw a line that’s parallel to none of these segments. If we move the line rightward through the field, keeping its slope constant, then it will encounter one point at a time, so we can stop it precisely midway through the field (or anywhere else we like).

February 26, 2012January 31, 2013 | Puzzles

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