The Two Poles

dudeney poles puzzle

A puzzle by Henry Dudeney:

A man planted two poles upright in level ground. One pole was six and a half feet and the other seven feet seven inches above ground. From the top of each pole he tied a string to the bottom of the other — just where it entered the ground. Now, what height above the ground was the point where the two strings crossed one another? The hasty reader will perhaps say, “You have forgotten to tell us how far the poles were apart.” But that point is of no consequence whatever, as it does not affect the answer!

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One pole was 78 inches high, and the other 91 inches. Multiply these two numbers together, and also add them together. Then divide the first result by the second, and you get 42 inches as the height above ground of the point where the two strings intersect, no matter how far the poles may be apart.

September 28, 2012February 7, 2013 | Puzzles

Black and White

By R.F. Bennett. White to mate in two moves.

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1. Bc7 and the queen will mate. From J. Paul Taylor, Chess Chips, 1878.

September 27, 2012February 7, 2013 | Puzzles

Out of Rank

beasley chess puzzle

A puzzle from John Beasley’s The Mathematics of Games (2006): Black has just moved. What is the smallest number of moves that can have been played in this game?

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Black can have moved only his knights and his rooks. The knights began on squares of different colors, and a moving knight lands on light and dark squares alternately, so between them the two knights have made an even number of moves (even if they’ve exchanged places). While the knights were abroad the rooks may have shifted back and forth a bit, but they’re on their home squares now and so have made an even number of moves themselves (possibly zero). So Black has made an even number of moves altogether. If Black has made an even number of moves, then so has White. But this presents a problem: White’s knights and rooks have moved an even number of times, but the f-pawn’s position on f3 means that White has made an additional move, giving an odd total. In order to make the total even we must find an odd quantity of additional moves that White can have made, and they can’t have been made with pawns, knights, or rooks. The simplest solution is that the king has made seven moves, e.g., e1-f2-e3-f4-g4-g3-f2-e1. So the answer is that each player has made at least eight moves: The shortest game that could produce the position above with White to play begins with 1. f3 and then the white king makes seven moves while Black marks time with his knights (and perhaps his rooks).

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September 25, 2012February 7, 2013 | Puzzles

Drinking Problem

I have a 16-ounce bottle of wine and want to make it last as long as possible, so I establish the following plan: On the first day I’ll drink 1 ounce of wine and refill the bottle with water. On the second day I’ll drink 2 ounces of the mixture and refill the bottle with water. On the third day I’ll drink three ounces of the mixture and again refill the bottle with water. If I continue until the bottle is empty, how many ounces of water will I have drunk?

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It’s easy to get bogged down in calculating proportions, but there’s a simpler solution. If I continue until the bottle is empty, then eventually I will have drunk all the water that I’ve added to the bottle. I add 1 ounce on the first day, 2 on the second, and so on to the 15th day. (I drink off the entire bottle on the 16th day, so no water is added then.) So in all I’ll have drunk 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 = 120 ounces of water. (Another solution is to calculate that in the end I’ve drunk 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 136 ounces in total, and then deduct the 16 ounces of wine.)

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September 23, 2012February 6, 2013 | Puzzles

Weighty Matters

I’ve accidentally turned the calibration dial on my bathroom scale, so its readings are skewed by a consistent amount. Apart from that it works fine, though. When I stand on the scale it reads 170 pounds, when my wife stands on it it reads 130, and when we stand on it together it reads 292 pounds. How should I adjust the scale?

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When I’m standing on the scale and my wife joins me, the difference in the readings will show us her true weight, 292 – 170 = 122 pounds. So the scale is set 130 – 122 = 8 pounds too high.

September 22, 2012February 6, 2013 | Puzzles

Black and White

From Alexander H. Robbins, A Collection of Chess Problems, 1887. White to mate in two moves.

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The quiet 1. Qb3 threatens six different mates.

September 21, 2012February 6, 2013 | Puzzles

Black and White

BCM chess problem

From the British Chess Magazine, January 1899. White to mate in two.

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After 1. Qb1, Black must accept one of six different mates: 1. … d5 2. Qb6# 1. … g6 2. Qb2# 1. … Ng6 2. Ne4# 1. … Nxg4 2. Nxg4# 1. … Nf5 2. Qxf5# 1. … Ng8 2. Qf5#

September 15, 2012February 6, 2013 | Puzzles

Waste Not …

A puzzle from Polish mathematician Paul Vaderlind:

John is swimming upstream in a river when he loses his goggles. He lets them go and continues upstream for 10 minutes, then decides to turn around and retrieve them. He catches up with them at a point one half mile from the point where he lost them. Is the river flowing faster than 1 mile per hour? (Assume that John swims at the same strength throughout.)

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Imagine an observer in a boat near the goggles. From her point of view, the goggles aren’t moving — John simply swims away from them for 10 minutes and will swim back in another 10 minutes (since he’s swimming at a constant speed). If the goggles have traveled half a mile in these 20 minutes, then the river is flowing at 1.5 miles per hour.

September 14, 2012February 6, 2013 | Puzzles

Twice Blessed

When Linus Pauling won his second Nobel Prize in 1962, he joked that receiving his second Nobel was less remarkable than receiving his first: The chance of anyone receiving his first Nobel Prize is one in several billion (the population of the world), while the chance of receiving his second is one in several hundred (the number of living people who have received one prize).

What’s wrong with this argument?

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Nobel prize winners are chosen independently of past awards, so if Pauling’s chance of winning one prize was (say) 1 in 3,000,000,000, then he’d have the same chance of winning the second, and his chance of winning two would be 1 in 3,000,000,000². (Actually they’re considerably better than that — not everyone on Earth is equally deserving of the Nobel prize.) If we knew that the Nobel committee would award the 1962 prize only to an existing laureate, then Pauling’s argument would hold. (From The Ivan Morris Puzzle Book, 1972.)

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September 13, 2012February 6, 2013 | Puzzles

Side Issue

A conundrum by Russian puzzle maven Boris Kordemsky:

A work train composed of a locomotive and five cars has just stopped at a railway station when word comes that a passenger train is approaching. The smaller train must make way for it to pass through, but the station has only one siding, and this will accommodate only three cars (or an engine and two cars). How can it arrange to let the passenger train through?

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The work train backs into the siding and uncouples its three rearmost cars. Then it moves forward a sufficient distance. The passenger train passes entirely through the station, backs into the siding, and couples these three cars. Then it moves forward again into the main railway, backs entirely through the station, and uncouples the three cars. The work train backs into the siding, which will now accommodate it, and the passenger train can move forward on its way. Now the work train can emerge from the siding and recouple the three waiting cars.

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September 12, 2012February 6, 2013 | Puzzles