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

Flying Fur

loyd cat and dog puzzle

Sam Loyd devised this puzzle for P.T. Barnum:

A trained cat and dog run a race, one hundred feet straight away and return. The dog leaps three feet at each bound and the cat but two, but then she makes three leaps to his two. Now, under those circumstances, what are the possible outcomes of the race?

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As we might have expected, it’s a trick question. It appears at first that the race will end in a tie, as the two animals are covering ground at the same rate. The trouble occurs at the far end of the course. The cat, which covers 2 feet at a bound, can turn around neatly at the halfway point. The dog, which covers 3 feet, cannot: It will arrive at the 99-foot mark and then must make an additional leap to get past the same point. “In all, the dog must make 68 leaps to go the distance. But it jumps only two-thirds as quickly as the cat, so that while the cat is making 100 leaps the dog cannot make quite 67.”

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September 11, 2012February 6, 2013 | Puzzles

Lattice Work

lattice work grid

Plot five points at random at the intersections of a coordinate grid. Between each pair of points a line segment can be drawn. Prove that the midpoint of at least one of these segments occurs at an intersection of grid lines.

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The coordinates of each point can be categorized by their parity — for example, (1, 1) is odd-odd and (4, 17) is even-odd. The midpoint of a line segment between two points will occur at a lattice point only if the two endpoints fall into the same category; for example, the two points above won’t produce such a midpoint, but (6, 8) and (14, 2) will because both are even-even. There are only four possible categories: even-even, odd-odd, even-odd, and odd-even. And because we are plotting five points, at least two points must fall into the same category. The midpoint between these two points will occur at a lattice point. From L.C. Larsen, Problem Solving Through Problems, 1983.

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September 10, 2012February 6, 2013 | Puzzles

Black and White

crane chess problem

By William Crane Jr., from the Sydney Town and Country Journal, 1877. White to mate in two moves.

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After the surprising 1. Bd3!, Black will be mated by 2. Qd1, 2. Bf6, or 2. Qxe4.

September 7, 2012February 6, 2013 | Puzzles