Both were wrong, and the lady was badly cheated. She only got half the quantity that would be contained in the large bundle, and therefore should have been charged half the original price. A circle with the circumference half that of another must have its area a quarter that of the other.

(Henry E. Dudeney, “The Paradox Party,” Strand 38:228 [January 1910], 626-632.)

There’s no legal move by which Black could have removed one of his own pieces from the board, so he must have moved the king. It’s illegal for two kings ever to occupy adjacent squares, so the king must have moved to a8 from a7. White’s previous move must have placed him in check there. White cannot have accomplished this with a bishop move, as the white bishop cannot have moved to its present position from another diagonal. The only other possibility is a discovered check by a knight:

So, in this position, White moved the knight from b6 to a8, discovering check, and Black captured the knight with his king.

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.)

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.

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.

Turn the board around.

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.

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.

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.

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.