A Tangled Tale

right. Of the 15, I may dismiss Alphabetical Phantom, Bog-Oak, Dinah Mite, Fifee, Galanthus Nivalis Major (I fear the cold spring has blighted our Snowdrop), Guy, H.M.S. Pinafore, Janet, and Valentine with the simple remark that they insist on the unfortunate lodgers keeping to the pavement. (I used the words “crossed to Number Seventy-three” for the special purpose of showing that shortcuts were possible.) Sea-Breeze does the same, and adds that “the result would be the same” even if they crossed the Square, but gives no proof of this. M. M. draws a diagram, and says that No. 9 is the house, “as the diagram shows.” I cannot see how it does so. Old Cat assumes that the house must be No. 9 or No. 73. She does not explain how she estimates the distances. Bee’s Arithmetic is faulty: she makes 169+442+130=741. (I suppose you mean 741, which would be a little nearer the truth. But roots cannot be added in this manner. Do you think 9+16 is 25, or even 25?) But Ayr’s state is more perilous still: she draws illogical conclusions with a frightful calmness. After pointing out (rightly) that AC is less than BD she says, “therefore the nearest house to the other three must be A or C.” And again, after pointing out (rightly) that B and D are both within the half-square containing A, she says “therefore” AB+AD must be less than BC+CD. (There is no logical force in either “therefore.” For the first, try Nos. 1, 21, 60, 70: this will make your premise true, and your conclusion false. Similarly, for the second, try Nos. 1, 30, 51, 71.)

Of the five partly-right solutions, Rags and Tatters and Mad Hatter (who send one answer between them) make No. 25 6 units from the corner instead of 5. Cheam, E. R. D. L., and Meggy Potts leave openings at the corners of the Square, which are not in the data: moreover Cheam gives values for the distances without any hint that they are only approximations. Crophi and Mophi make the bold and unfounded assumption that there were really 21 houses on each side, instead of 20 as stated by Balbus. “We may assume,” they add, “that the doors of Nos. 21, 42, 63, 84, are invisible from the centre of the Square”! What is there, I wonder, that Crophi and Mophi would not assume?

Of the five who are wholly right, I think Bradshaw Of the Future, Caius, Clifton C., and Martreb deserve special praise for their full analytical solutions. Matthew Matticks picks out No. 9, and proves it to be the right house in two ways, very neatly and ingeniously, but why he picks it out does not appear. It is an excellent synthetical proof, but lacks the analysis which the other four supply.

Class List.

1. • Bradshaw of the Future.

   • Caius.

   • Clifton C.

   • Martreb.

2. • Matthew Matticks.

3. • Cheam.

   • Crophi and Mophi.

   • E. R. D. L.

   • Meggy Potts.

   • Rags and Tatters.

   • Mad Hatter.

A remonstrance has reached me from Scrutator on the subject of Knot I, which he declares was “no problem at all.” “Two questions,” he says, “are put. To solve one there is no data: the other answers itself.” As to the first point, Scrutator is mistaken; there are (not “is”) data sufficient to answer the question. As to the other, it is interesting to know that the question “answers itself,” and I am sure it does the question great credit: still I fear I cannot enter it on the list of winners, as this competition is only open to human beings.

Answers to Knot III

Problem.⁠—(1) “Two travellers, starting at the same time, went opposite ways round a circular railway. Trains start each way every 15 minutes, the easterly ones going round in 3 hours, the westerly in 2. How many trains did each meet on the way, not counting trains met at the terminus itself?” (2) “They went round, as before, each traveller counting as ‘one’ the train containing the other traveller. How many did each meet?”

Answers.⁠—(1) 19. (2) The easterly traveller met 12; the other 8.

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The trains one way took 180 minutes, the other way 120. Let us take the L. C. M., 360, and divide the railway into 360 units. Then one set of trains went at the rate of 2 units a minute and at intervals of 30 units; the other at the rate of 3 units a minute and at intervals of 45 units. An easterly train starting has 45 units between it and the first train it will meet: it does ⅖ths of this while the other does ⅗ths, and thus meets it at the end of 18 units, and so all the way round. A westerly train starting has 30 units between it and the first train it will meet: it does ⅗ths of this while the other does ⅖ths, and thus meets it at the end of 18 units, and so all the way round. Hence if the railway be divided, by 19 posts, into 20 parts, each containing 18 units, trains meet at every post, and, in (1), each traveller passes 19 posts in going round, and so meets 19 trains. But, in (2), the easterly traveller only begins to count after traversing ⅖ths of the journey, i.e., on reaching the 8th post, and so counts 12 posts: similarly the other counts 8. They meet at the end of ⅖ths of 3 hours, or ⅗ths of 2 hours, i.e., 72 minutes.

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Forty-five answers have been received. Of these 12 are beyond the reach of discussion, as they give no working. I can but enumerate their names. Ardmore, E. A., F. A. D., L. D., Matthew Matticks, M. E. T., Poo-Poo, and The Red Queen are all wrong. Beta and Rowena have got (1) right and (2) wrong. Cheeky Bob and Nairam give the right answers,