Black and White

charosh chess puzzle

By M. Charosh, from the Fairy Chess Review, 1937. White to mate in zero moves.

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Turn the board around.

October 19, 2012February 7, 2013 | Puzzles

The Lodging-House Difficulty

A puzzle by Henry Dudeney:

dudeney lodging-house difficulty

The Dobsons secured apartments at Slocomb-on-Sea. There were six rooms on the same floor, all communicating, as shown in the diagram. The rooms they took were numbers 4, 5, and 6, all facing the sea.

But a little difficulty arose. Mr. Dobson insisted that the piano and the bookcase should change rooms. This was wily, for the Dobsons were not musical, but they wanted to prevent any one else playing the instrument.

Now, the rooms were very small and the pieces of furniture indicated were very big, so that no two of these articles could be got into any room at the same time. How was the exchange to be made with the least possible labour? Suppose, for example, you first move the wardrobe into No. 2; then you can move the bookcase to No. 5 and the piano to No. 6, and so on.

It is a fascinating puzzle, but the landlady had reasons for not appreciating it. Try to solve her difficulty in the fewest possible removals with counters on a sheet of paper.

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The shortest possible way is to move the articles in the following order: piano, bookcase, wardrobe, piano, cabinet, chest of drawers, piano, wardrobe, bookcase, cabinet, wardrobe, piano, chest of drawers, wardrobe, cabinet, bookcase, piano. Thus seventeen removals are necessary. The landlady could the move chest of drawers, wardrobe, and cabinet. Mr. Dobson did not mind the wardrobe and chest of drawers changing rooms so long as he secured the piano.

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October 16, 2012February 7, 2013 | Puzzles

Seven Tails

Here are seven pennies, all heads up. In a single move you can turn over any four of them. By repeatedly making such moves, can you eventually turn all seven pennies tails up?

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No, it can’t be done. On a given move, suppose that t pennies change from tails to heads. That means that 4-t pennies change from heads to tails. So this move has changed the total number of tails by (4-t) – t = 4-2t = 2(2-t). Since this number is a multiple of 2, it will always be even. If we started with 0 tails, and each move changes the total number of tails by an even number, then we can never reach 7 tails. From Paul Vaderlind, Richard K. Guy, and Loren C. Larson, The Inquisitive Problem Solver, 2002.

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October 13, 2012February 7, 2013 | Puzzles

Black and White

By Francis Healey. White to mate in two moves.

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White wins with the preposterous-looking 1. Qh7!, threatening 2. Qxd7#. If Black captures the queen then White mates with the knight. From A Collection of Two Hundred Chess Problems, 1866.

October 12, 2012February 7, 2013 | Puzzles

Opposites Exact

Prove that, at any given moment, there are two points on the equator that are diametrically opposed yet have the same temperature.

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Call two opposing points A and B, and suppose that the temperature at A is higher than B. So A – B is positive. Now rotate both points around the equator, maintaining their opposition. Their difference can’t remain positive, because we know that when they’ve traded places it will be negative (the opposite of its initial value). Because temperature varies continuously, there must be some position in between where their difference is zero.

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October 11, 2012February 7, 2013 | Puzzles

Black and White

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

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1. Qg6 and White will mate with 2. Qb1, 2. Rb1, 2. Na7, or 2. Rxc5. From J. Paul Taylor, Chess Chips, 1878.

October 5, 2012February 7, 2013 | Puzzles

Injured List

Another puzzle from Henry Dudeney:

“It is a glorious game!” an enthusiast was heard to exclaim. “At the close of last season, of the footballers of my acquaintance, four had broken their left arm, five had broken their right arm, two had the right arm sound, and three had sound left arms.” Can you discover from that statement what is the smallest number of players that the speaker could be acquainted with?

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The smallest possible number of men is seven. They could be accounted for in three different ways. (1) Two with both arms sound, one with broken right arm, and four with both arms broken. (2) One with both arms sound, one with broken left arm, two with broken right arm, and three with both arms broken. (3) Two with left arm broken, three with right arm broken, and two with both arms broken. But if every man was injured, the last case is the only one that would apply.

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October 2, 2012February 7, 2013 | Puzzles

The Dwarfs Problem

From the 1977 all-Soviet-Union Mathematical Olympiad:

Seven dwarfs are sitting at a round table. Each has a cup, and some cups contain milk. Each dwarf in turn pours all his milk into the other six cups, dividing it equally among them. After the seventh dwarf has done this, they find that each cup again contains its initial quantity of milk. How much milk does each cup contain, if there were 42 ounces of milk altogether?

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Since the final state matches the initial state, we can imagine this process going on continuously. Consider the dwarf whose cup contains the smallest amount of milk just before he begins pouring. Call him D, and call this quantity of milk a ounces. Since D’s cup contains the least milk before pouring, then each other dwarf has an equal or greater amount in his own cup before pouring, and thus gives D at least a/6 ounces. This would leave D with at least a ounces in his own cup just before pouring; the only way he can receive precisely a ounces, as we know he does, is if each other dwarf gives him precisely a/6 ounces and no more, i.e., that each cup, just before pouring, contains the same quantity of milk. So after any given pouring the cups contain a, 5/6a, 4/6a, 3/6a, 2/6a, 1/6a, and 0 ounces; if there are 42 ounces in total then this works out to 12, 10, 8, 6, 4, 2, and 0 ounces.

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October 1, 2012February 7, 2013 | Puzzles

Tiling Task

tiling task 1

We’ve removed two squares from this 7×8 grid, so that it numbers 54 squares. Can it be covered orthogonally with tiles like the one at right, each of which covers exactly three squares?

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No. Number the squares as shown. Now each tile, however it’s placed, must cover a 1, a 2, and a 3. But the grid contains 19 1s, 18 2s, and 17 3s. So the task is impossible. From Pierre Berloquin, The Garden of the Sphinx, 1981.

September 30, 2012February 7, 2013 | Puzzles

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