Lazy Tennis

You’ve just won a set of singles tennis. What’s the least number of times your racket can have struck the ball? Remember that if you miss the ball while serving, it’s a fault.

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Once. You swing and miss each time you serve, and your opponent commits a double fault each time he serves. Eventually you find yourself serving at 6-5 or 7-6 in the tiebreak, and then you heroically serve an ace. From Dick Hess, All-Star Mathlete Puzzles, 2009. UPDATE: A reader points out that a superlatively lazy player can win a set without ever hitting the ball, thanks to the hindrance rule: 26. HINDRANCE If a player is hindered in playing the point by a deliberate act of the opponent(s), the player shall win the point. “On the crucial tie-break point, your opponent dashes across the net and tackles you, awarding you the point and the set.” (Thanks, Scott.)

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October 24, 2012February 7, 2013 | Puzzles

Twisted Math

The Renaissance mathematician Niccolò Tartaglia would use this bewildering riddle to assess neophytes in logic:

If half of 5 were 3, what would a third of 10 be?

What’s the answer?

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The question asks us to discover whatever strange factor could cause 5/2 to give the result 3, and to apply the same factor to 10/3. So: 5/2 : 3 = 10/3 : x Solving for x gives the result 4.

October 22, 2012February 7, 2013 | Puzzles

Stride Right

A mother takes two strides to her daughter’s three. If they set out walking together, each starting with the right foot, when will they first step together with the left?

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Never. After the mother’s first four strides and the daughter’s first six, they will again step together with the right foot, and the cycle repeats. Nowhere in that interval do they step together with the left foot.

October 21, 2012February 7, 2013 | Puzzles

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