A problem by Russian mathematician Vyacheslav Proizvolov:

At a party each girl danced with three boys, and each boy danced with three girls. Prove that the number of girls at the party was equal to the number of boys.

A problem by Russian mathematician Vyacheslav Proizvolov:

At a party each girl danced with three boys, and each boy danced with three girls. Prove that the number of girls at the party was equal to the number of boys.

A problem by Raymond Smullyan. The diagram above shows the final position in a chess game in which nothing has moved from a white square to a black one or vice versa. One piece has been omitted from the diagram. What color is the square that it stands on?

A puzzle by Henry Dudeney:

When visiting with a friend one of our hospitals for wounded soldiers, I was informed that exactly two-thirds of the men had lost an eye, three-fourths had lost an arm, and four-fifths had lost a leg. ‘Then,’ I remarked to my friend, ‘it follows that at least twenty-six of the men must have lost all three — an eye, an arm, and a leg.’ That being so, can you say exactly how many men were in the hospital? It is a very simple calculation, but I have no doubt it will perplex a good many readers.

A man goes into a 7-11 store, buys four items, and notices that the bill totals $7.11. Even more interestingly, the product of the four prices is 7.11. What are the prices?

The answer is $1.20, $1.25, $1.50, and $3.16. There’s no thunderbolt insight to find; the problem yields to patient consideration.

Who came up with it? The most common credit I’ve seen is Doug Brumbaugh of the University of Central Florida. I found it in *Crux Mathematicorum*; see problem M203 in the September 2006 issue for Richard K. Guy’s solution.

A puzzle by V. Dubrovsky, from *Quantum*, January-February 1992:

In a certain planetary system, no two planets are separated by the same distance. On each planet sits an astronomer who observes the planet closest to hers. Prove that if the total number of planets is odd, there must be a planet that no one is observing.

Arthur and Robert are identical twins. One always lies, and the other always tells the truth, but you don’t know which is the liar. One day you meet one of them and want to find out whether it’s Arthur or Robert. But you can ask only one yes/no question, and the question can’t contain more than three words. What question will do? Alternatively, suppose you want to find out whether it’s Arthur or Robert who’s truthful. What three-word yes/no question will reveal the answer?

By W. Timbrell Pierce. White to mate in two moves.

A clever conundrum from the U.K. Government Communications Headquarters Christmas Puzzle Quiz — it doesn’t even appear to be a question:

42°15′N 72°15′W, 53°52′N 44°50′E, 37°49′N 85°29′W, 39°37′N 75°56′W, 40°57′N 40°17′E, 51°54′N 02°04′W?

The answer is 52°12′N 1°41′W. The listed coordinates identify Ware, Issa, Bardstown, North-East, Of, and Cheltenham. Where is a Bard’s town northeast of Cheltenham? Stratford-upon-Avon! Its own coordinates make up the answer.

GCHQ is a great fount of excellent original puzzles — some are given on their website, they’ve published two collections, and they regularly post brainteasers on Twitter.

From Lee Sallows:

Earlier this year, Futility Closet featured a puzzle based upon the well-known 7-segment display. Less well known is the 15-cell display shown in Figure 1, in which each decimal digit appears as a pattern of highlighted cells within a 3×5 rectangle. Call these the

smallrectangles. Observe also that the digit 1 is represented as the vertical column of 5 cellsin the centreof its small rectangle rather than as either of the two alternative columns immediately to left and right, a detail that is important in view of what follows.Figure 1.

Figure 2 shows a pair of readouts using 15-cell displays each arranged in the form of a (large) 3×5 rectangle that mirrors the smaller rectangles just mentioned. The two readouts describe each other. The top left cell in the right-hand readout contains the number 18. A check will show that the number of highlighted top left cells appearing in the left-hand readout is indeed 18. Take for instance the top left-hand cell in the left-hand readout. It contains the number 17, which employs two digits, 1 and 7. None of the 5 cells forming the digit 1 is in top left position within its small rectangle. But the leftmost cell in digit 7 is indeed in top left position. Proceeding next to the left-hand readout’s top centre cell we find two cells in top left position: one in digit 2 and one in digit 4. The score of top left cells so far is thus 1+2 = 3. Continuing in normal reading order, a list of the left-hand readout numbers followed by their top left cell scores in brackets is as follows: 17 (1), 24 (2), 17(1), 13(1), 9(1), 15(1), 17(1), 25(2), 17(1), 8(1), 9(1), 14(1), 15(1), 24(2), 17(1). The sum of the scores is 18, as predicted.

In the same way, a number occupying position

xin either of the readouts will be found to identify the total number of cells occurring in positionxwithin the digits of the other readout. That is, the two readouts are co-descriptive, they describe each other.Figure 2.

Recalling now the solution to the earlier mentioned 7-segment display puzzle, some readers may recall that it involved an iterative process that terminated in a loop of length 4. Likewise, the pair of readouts in Figure 2 are the result of a similar process, but now terminating in a loop of length 2. In that case we were counting segments, here we are counting cells. An obvious question thus prompted is: What kind of a readout would result from a loop of length 1? The answer is simple: a description of the readout resulting from a loop of length 1 would be a copy of the same readout. That is, it will be a self-descriptive readout, the description of which is identical to itself. Such a readout does indeed exist. Can the reader find it?

A brainteaser by Y. Bogaturov, via *Quantum*, November-December 1991:

In square *ABCD*, point *L* divides diagonal *AC* in the ratio 3:1 and *K* is the midpoint of side *AB*. Prove that angle *KLD* is a right angle.