“Counting the Wounded”

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.

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“The three fractions are respectively 40/60, 45/60, and 48/60. Add together 40, 45, and 48, and deduct twice 60. The result is 13, as the minimum number for every 60 patients. Therefore as the minimum (who could have each lost an eye, an arm, and a leg) was 26, the number of patients must have been 120.” From Modern Puzzles and How to Solve Them, 1926.

August 2, 2022August 1, 2022 | Puzzles

The 7-11 Problem

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.

July 29, 2022July 28, 2022 | Puzzles

Scrutiny

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.

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Of the n planets in the system, consider the two that are closest together. These two astronomers must be looking at each other. If any of the remaining n – 2 astronomers are looking at one of these planets, then there won’t be enough astronomers left to observe all the remaining n – 2 planets. On the other hand, if none of the remaining astronomers are looking at the two closest-set planets, then we can discard those two and ask our original question of the n – 2 planets that remain: Which two are closest together? Again, those two astronomers must be observing each other. And so on. Each time we discard a pair of planets we’ll have some odd number remaining. But that number can’t decrease forever (it can’t be negative). We must eventually arrive at 1, a planet that no astronomer is observing.

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July 22, 2022July 21, 2022 | Puzzles

Double Trouble

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?

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Raymond Smullyan points out that there’s a pleasing symmetry between the answers to these problems. To find out which twin is Arthur, you can ask, “Is Arthur truthful?” To find out whether Arthur is truthful, ask, “Are you Arthur?” (From Smullyan’s 2009 book Logical Labyrinths, via Jason Rosenhouse’s Games for Your Mind, 2020.)

July 15, 2022July 14, 2022 | Puzzles

Black and White

pierce chess problem

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

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White wins with the quiet 1. Ne2! From Chess Problems, 1873.

July 14, 2022July 13, 2022 | Puzzles

A Matter of Degree

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.

July 7, 2022July 6, 2022 | Puzzles

Readouts Revisited

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 small rectangles. Observe also that the digit 1 is represented as the vertical column of 5 cells in the centre of 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 x in either of the readouts will be found to identify the total number of cells occurring in position x within 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?

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The figure below shows one of two solutions to the puzzle. As the reader can verify, the position of each number in the readout is the position of those yellow cells within their smaller rectangles that the number counts. Consider now the digit 6 that occurs in the leftmost cell of the fourth row of the readout. If we excise the yellow cell in that same position within the digit it will be turned from a 6 into a 5. Which will still be an accurate count of the yellow cells occupying that position. So changing the 6 into a 5 does not affect the readout’s self-descriptive property, and thus provides us with a genuine, if trivially distinct, second solution. Figure 3. (Thanks, Lee!)

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July 5, 2022July 4, 2022 | Puzzles

Tilt

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.

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If we divide the square into 16 smaller squares, as shown, K and L fall at nodes in the resulting grid. Now rotating the diagram through 90° around L carries triangle LMD into triangle LNK. So angle DLK must equal the angle of rotation, or 90°.

July 2, 2022July 1, 2022 | Puzzles

Three for Three

A problem from the February 2006 issue of Crux Mathematicorum:

In a set of five positive integers, show that it’s always possible to choose three such that their sum is a multiple of 3.

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This solution is by Geneviève Lalonde. Any integer must fall into one of three groups: It’s a multiple of 3. It’s one less than a multiple of 3. It’s one more than a multiple of 3. In our set of five numbers, either it’s the case that all three of these groups are represented or it’s not. If all three are represented, then the set includes numbers of the form 3a – 1, 3b, and 3c + 1. The sum of these is 3a – 1 + 3b + 3c + 1 = 3(a + b + c), which is a multiple of 3. If not all three groups are represented, then one of the groups must contain at least three numbers from our set of five. If there are three numbers in the first group, they’ll have the form 3a, 3b, and 3c. If there are three numbers in the second group, they’ll have the form 3a – 1, 3b – 1, and 3c – 1. If there are three numbers in the third group, they’ll have the form 3a + 1, 3b + 1, and 3c + 1. In each case, the sum of the three numbers is a multiple of 3.

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