I have received several letters on the subjects of Knots II. and VI., which lead me to think some further explanation desirable.

In Knot II., I had intended the numbering of the houses to begin at one corner of the Square, and this was assumed by most, if not all, of the competitors. Trojanus however says "assuming, in default of any information, that the street enters the square in the middle of each side, it may be supposed that the numbering begins at a street." But surely the other is the more natural assumption?

In Knot VI., the first Problem was of course a mere *jeu de mots*, whose presence I thought excusable in a series of Problems whose aim is to entertain rather than to instruct: but it has not escaped the contemptuous criticisms of two of my correspondents, who seem to think that Apollo is in duty bound to keep his bow always on the stretch. Neither of them has guessed it: and this is true human nature. Only the other day—the 31st of September, to be quite exact—I met my old friend Brown, and gave him a riddle I had just heard. With one great effort of his colossal mind, Brown guessed it. "Right!" said I. "Ah," said he, "it’s very neat—very neat. And it isn’t an answer that would occur to everybody. Very neat indeed." A few yards further on, I fell in with Smith and to him I propounded the same riddle. He frowned over it for a minute, and then gave it up. Meekly I faltered out the answer. "A poor thing, sir!" Smith growled, as he turned away. "A very poor thing! I wonder you care to repeat such rubbish!" Yet Smith’s mind is, if possible, even more colossal than Brown’s.

The second Problem of Knot VI. is an example in ordinary Double Rule of Three, whose essential feature is that the result depends on the variation of several elements, which are so related to it that, if all but one be constant, it varies as that one: hence, if none be constant, it varies as their product. Thus, for example, the cubical contents of a rectangular tank vary as its length, if breadth and depth be constant, and so on; hence, if none be constant, it varies as the product of the length, breadth, and depth.

When the result is not thus connected with the varying elements, the Problem ceases to be Double Rule of Three and often becomes one of great complexity.

To illustrate this, let us take two candidates for a prize, *A* and *B*, who are to compete in French, German, and Italian:

(*a*) Let it be laid down that the result is to depend on their *relative* knowledge of each subject, so that, whether their marks, for French, be "1, 2" or "100, 200," the result will be the same: and let it also be laid down that, if they get equal marks on 2 papers, the final marks are to have the same ratio as those of the 3rd paper. This is a case of ordinary Double Rule of Three. We multiply *A*’s 3 marks together, and do the same for *B*. Note that, if *A* gets a single "0," his final mark is "0," even if he gets full marks for 2 papers while *B* gets only one mark for each paper. This of course would be very unfair on *A*, though a correct solution under the given conditions.

(*b*) The result is to depend, as before, on *relative* knowledge; but French is to have twice as much weight as German or Italian. This is an unusual form of question. I should be inclined to say "the resulting ratio is to be nearer to the French ratio than if we multiplied as in (*a*), and so much nearer that it would be necessary to use the other multipliers *twice* to produce the same result as in (*a*):" *e.g.* if the French Ratio were ^{2}⁄_{10}, and the others ^{2}⁄_{9}, ^{1}⁄_{9} so that the ultimate ratio, by method (*a*), would be ^{2}⁄_{45}, I should multiply instead by ^{2}⁄_{3}, ^{1}⁄_{3}, giving the result, ^{1}⁄_{3} which is nearer to ^{2}⁄_{10} than if he had used method (*a*).

(*c*) The result is to depend on *actual* amount of knowledge of the 3 subjects collectively. Here we have to ask two questions. (1) What is to be the "unit" (*i.e.* "standard to measure by") in each subject? (2) Are these units to be of equal, or unequal value? The usual "unit" is the knowledge shown by answering the whole paper correctly; calling this "100," all lower amounts are represented by numbers between "0" and "100." Then, if these units are to be of equal value, we simply add *A*’s 3 marks together, and do the same for *B*.

(*d*) The conditions are the same as (*c*), but French is to have double weight. Here we simply double the French marks, and add as before.

(*e*) French is to have such weight, that, if other marks be equal, the ultimate ratio is to be that of the French paper, so that a "0" in this would swamp the candidate: but the other two subjects are only to affect the result collectively, by the amount of knowledge shown, the two being reckoned of equal value. Here I should add *A*’s German and Italian marks together, and multiply by his French mark.

But I need not go on: the problem may evidently be set with many varying conditions, each requiring its own method of solution. The Problem in Knot VI. was meant to belong to variety (*a*), and to make this clear, I inserted the following passage:

"Usually the competitors differ in one point only. Thus, last year, Fifi and Gogo made the same number of scarves in the trial week, and they were equally light; but Fifi’s were twice as warm as Gogo’s, and she was pronounced twice as good."

What I have said will suffice, I hope, as an answer to Balbus, who holds that (*a*) and (*c*) are the only possible varieties of the problem, and that to say "We cannot use addition, therefore we must be intended to use multiplication," is "no more illogical than, from knowledge that one was not born in the night, to infer that he was born in the daytime"; and also to Fifee, who says "I think a little more consideration will show you that our ’error of *adding* the proportional numbers together for each candidate instead of *multiplying*’ is no error at all." Why, even if addition *had* been the right method to use, not one of the writers (I speak from memory) showed any consciousness of the necessity of fixing a "unit" for each subject. "No error at all!" They were positively steeped in error!

One correspondent (I do not name him, as the communication is not quite friendly in tone) writes thus:—"I wish to add, very respectfully, that I think it would be in better taste if you were to abstain from the very trenchant expressions which you are accustomed to indulge in when criticising the answer. That such a tone must not be" ("be not"?) "agreeable to the persons concerned who have made mistakes may possibly have no great weight with you, but I hope you will feel that it would be as well not to employ it, *unless you are quite certain of being correct yourself*." The only instances the writer gives of the "trenchant expressions" are "hapless" and "malefactors." I beg to assure him (and any others who may need the assurance: I trust there are none) that all such words have been used in jest, and with no idea that they could possibly annoy any one, and that I sincerely regret any annoyance I may have thus inadvertently given. May I hope that in future they will recognise the distinction between severe language used in sober earnest, and the "words of unmeant bitterness," which Coleridge has alluded to in that lovely passage beginning "A little child, a limber elf"? If the writer will refer to that passage, or to the preface to "Fire, Famine, and Slaughter," he will find the distinction, for which I plead, far better drawn out than I could hope to do in any words of mine.

The writer’s insinuation that I care not how much annoyance I give to my readers I think it best to pass over in silence; but to his concluding remark I must entirely demur. I hold that to use language likely to annoy any of my correspondents would not be in the least justified by the plea that I was "quite certain of being correct." I trust that the knot-untiers and I are not on such terms as those!

I beg to thank *G. B.* for the offer of a puzzle—which, however, is too like the old one "Make four 9’s into 100."

§ 1. The Pigs.

*Problem.*—Place twenty-four pigs in four sties so that, as you go round and round, you may always find the number in each sty nearer to ten than the number in the last.

*Answer.*—Place 8 pigs in the first sty, 10 in the second, nothing in the third, and 6 in the fourth: 10 is nearer ten than 8; nothing is nearer ten than 10; 6 is nearer ten than nothing; and 8 is nearer ten than 6.

This problem is noticed by only two correspondents. Balbus says "it certainly cannot be solved mathematically, nor do I see how to solve it by any verbal quibble." Nolens Volens makes Her Radiancy change the direction of going round; and even then is obliged to add "the pigs must be carried in front of her"!

§ 2. The Grurmstipths.

*Problem.*—Omnibuses start from a certain point, both ways, every 15 minutes. A traveller, starting on foot along with one of them, meets one in 12½ minutes: when will he be overtaken by one?

*Answer.*—In 6¼ minutes.

*Solution.*—Let "*a*" be the distance an omnibus goes in 15 minutes, and "*x*" the distance from the starting-point to where the traveller is overtaken. Since the omnibus met is due at the starting-point in 2½ minutes, it goes in that time as far as the traveller walks in 12½; *i.e.* it goes 5 times as fast. Now the overtaking omnibus is "*a*" behind the traveller when he starts, and therefore goes "*a* + *x*" while he goes "*x*." Hence *a* + *x* = 5*x*; *i.e.* 4*x* = *a*, and *x* = *a*/4. This distance would be traversed by an omnibus in ^{15}⁄_{4} minutes, and therefore by the traveller in 5 × ^{15}⁄_{4}. Hence he is overtaken in 18¾ minutes after starting, *i.e.* in 6¼ minutes after meeting the omnibus.

Four answers have been received, of which two are wrong. Dinah Mite rightly states that the overtaking omnibus reached the point where they met the other omnibus 5 minutes after they left, but wrongly concludes that, going 5 times as fast, it would overtake them in another minute. The travellers are 5-minutes-walk ahead of the omnibus, and must walk 1-4th of this distance farther before the omnibus overtakes them, which will be 1-5th of the distance traversed by the omnibus in the same time: this will require 1¼ minutes more. Nolens Volens tries it by a process like "Achilles and the Tortoise." He rightly states that, when the overtaking omnibus leaves the gate, the travellers are 1-5th of "*a*" ahead, and that it will take the omnibus 3 minutes to traverse this distance; "during which time" the travellers, he tells us, go 1-15th of "*a*" (this should be 1-25th). The travellers being now 1-15th of "*a*" ahead, he concludes that the work remaining to be done is for the travellers to go 1-60th of "*a*," while the omnibus goes l-12th. The *principle* is correct, and might have been applied earlier.

CLASS LIST.

I.

Balbus.

Delta.

§ 1. The Buckets.

*Problem.*—Lardner states that a solid, immersed in a fluid, displaces an amount equal to itself in bulk. How can this be true of a small bucket floating in a larger one?

*Solution.*—Lardner means, by "displaces," "occupies a space which might be filled with water without any change in the surroundings." If the portion of the floating bucket, which is above the water, could be annihilated, and the rest of it transformed into water, the surrounding water would not change its position: which agrees with Lardner’s statement.

Five answers have been received, none of which explains the difficulty arising from the well-known fact that a floating body is the same weight as the displaced fluid. Hecla says that "only that portion of the smaller bucket which descends below the original level of the water can be properly said to be immersed, and only an equal bulk of water is displaced." Hence, according to Hecla, a solid, whose weight was equal to that of an equal bulk of water, would not float till the whole of it was below "the original level" of the water: but, as a matter of fact, it would float as soon as it was all under water. Magpie says the fallacy is "the assumption that one body can displace another from a place where it isn’t," and that Lardner’s assertion is incorrect, except when the containing vessel "was originally full to the brim." But the question of floating depends on the present state of things, not on past history. Old King Cole takes the same view as Hecla. Tympanum and Vindex assume that "displaced" means "raised above its original level," and merely explain how it comes to pass that the water, so raised, is less in bulk than the immersed portion of bucket, and thus land themselves—or rather set themselves floating—in the same boat as Hecla.

I regret that there is no Class-list to publish for this Problem.

§ 2. Balbus’ Essay.

*Problem.*—Balbus states that if a certain solid be immersed in a certain vessel of water, the water will rise through a series of distances, two inches, one inch, half an inch, &c., which series has no end. He concludes that the water will rise without limit. Is this true?

*Solution.*—No. This series can never reach 4 inches, since, however many terms we take, we are always short of 4 inches by an amount equal to the last term taken.

Three answers have been received—but only two seem to me worthy of honours.

Tympanum says that the statement about the stick "is merely a blind, to which the old answer may well be applied, *solvitur ambulando*, or rather *mergendo*." I trust Tympanum will not test this in his own person, by taking the place of the man in Balbus’ Essay! He would infallibly be drowned.

Old King Cole rightly points out that the series, 2, 1, &c., is a decreasing Geometrical Progression: while Vindex rightly identifies the fallacy as that of "Achilles and the Tortoise."

CLASS LIST.

I.

Old King Cole.

Vindex.

§ 3. The Garden.

*Problem.*—An oblong garden, half a yard longer than wide, consists entirely of a gravel-walk, spirally arranged, a yard wide and 3,630 yards long. Find the dimensions of the garden.

*Answer.*—60, 60½.

*Solution.*—The number of yards and fractions of a yard traversed in walking along a straight piece of walk, is evidently the same as the number of square-yards and fractions of a square-yard, contained in that piece of walk: and the distance, traversed in passing through a square-yard at a corner, is evidently a yard. Hence the area of the garden is 3,630 square-yards: *i.e.*, if *x* be the width, *x* (*x* + ½) = 3,630. Solving this Quadratic, we find *x* = 60. Hence the dimensions are 60, 60½.

Twelve answers have been received—seven right and five wrong.

C. G. L., Nabob, Old Crow, and Tympanum assume that the number of yards in the length of the path is equal to the number of square-yards in the garden. This is true, but should have been proved. But each is guilty of darker deeds. C. G. L.’s "working" consists of dividing 3,630 by 60. Whence came this divisor, oh Segiel? Divination? Or was it a dream? I fear this solution is worth nothing. Old Crow’s is shorter, and so (if possible) worth rather less. He says the answer "is at once seen to be 60 × 60½"! Nabob’s calculation is short, but "as rich as a Nabob" in error. He says that the square root of 3,630, multiplied by 2, equals the length plus the breadth. That is 60.25 × 2 = 120½. His first assertion is only true of a *square* garden. His second is irrelevant, since 60.25 is *not* the square-root of 3,630! Nay, Bob, this will *not* do! Tympanum says that, by extracting the square-root of 3,630, we get 60 yards with a remainder of 30/60, or half-a-yard, which we add so as to make the oblong 60 × 60½. This is very terrible: but worse remains behind. Tympanum proceeds thus:—"But why should there be the half-yard at all? Because without it there would be no space at all for flowers. By means of it, we find reserved in the very centre a small plot of ground, two yards long by half-a-yard wide, the only space not occupied by walk." But Balbus expressly said that the walk "used up the whole of the area." Oh, Tympanum! My tympa is exhausted: my brain is num! I can say no more.

Hecla indulges, again and again, in that most fatal of all habits in computation—the making *two* mistakes which cancel each other. She takes *x* as the width of the garden, in yards, and *x* + ½ as its length, and makes her first "coil" the sum of *x*½, *x*½, *x*-1, *x*-1, *i.e.* 4*x*-3: but the fourth term should be *x*-1½, so that her first coil is ½ a yard too long. Her second coil is the sum of *x*-2½, *x*-2½, *x*-3, *x*-3: here the first term should be *x*-2 and the last *x*-3½: these two mistakes cancel, and this coil is therefore right. And the same thing is true of every other coil but the last, which needs an extra half-yard to reach the *end* of the path: and this exactly balances the mistake in the first coil. Thus the sum total of the coils comes right though the working is all wrong.

Of the seven who are right, Dinah Mite, Janet, Magpie, and Taffy make the same assumption as C. G. L. and Co. They then solve by a Quadratic. Magpie also tries it by Arithmetical Progression, but fails to notice that the first and last "coils" have special values.

Alumnus Etonæ attempts to prove what C. G. L. assumes by a particular instance, taking a garden 6 by 5½. He ought to have proved it generally: what is true of one number is not always true of others. Old King Cole solves it by an Arithmetical Progression. It is right, but too lengthy to be worth as much as a Quadratic.

Vindex proves it very neatly, by pointing out that a yard of walk measured along the middle represents a square yard of garden, "whether we consider the straight stretches of walk or the square yards at the angles, in which the middle line goes half a yard in one direction and then turns a right angle and goes half a yard in another direction."

CLASS LIST.

I.

Vindex.

II.

Alumnus Etonæ.

Old King Cole.

III.

Dinah Mite.

Janet.

Magpie.

Taffy.