*Problem 1.*—*A* and *B* began the year with only 1,000*l.* a-piece. They borrowed nought; they stole nought. On the next New-Year’s Day they had 60,000*l.* between them. How did they do it?

*Solution.*—They went that day to the Bank of England. *A* stood in front of it, while *B* went round and stood behind it.

Two answers have been received, both worthy of much honour. Addlepate makes them borrow "0" and steal "0," and uses both cyphers by putting them at the right-hand end of the 1,000*l.*, thus producing 100,000*l.*, which is well over the mark. But (or to express it in Latin) At Spes infracta has solved it even more ingeniously: with the first cypher she turns the "1" of the 1,000*l.* into a "9," and adds the result to the original sum, thus getting 10,000*l.*: and in this, by means of the other "0," she turns the "1" into a "6," thus hitting the exact 60,000*l.*

CLASS LIST

I.

At Spes Infracta.

II.

Addlepate.

*Problem 2.*—*L* makes 5 scarves, while *M* makes 2: *Z* makes 4 while *L* makes 3. Five scarves of *Z*’s weigh one of *L*’s; 5 of *M*’s weigh 3 of *Z*’s. One of *M*’s is as warm as 4 of *Z*’s: and one of *L*’s as warm as 3 of *M*’s. Which is best, giving equal weight in the result to rapidity of work, lightness, and warmth?

*Answer.*—The order is *M*, *L*, *Z*.

*Solution.*—As to rapidity (other things being constant) *L*’s merit is to *M*’s in the ratio of 5 to 2: *Z*’s to *L*’s in the ratio of 4 to 3. In order to get one set of 3 numbers fulfilling these conditions, it is perhaps simplest to take the one that occurs *twice* as unity, and reduce the others to fractions: this gives, for *L*, *M*, and *Z*, the marks 1, ^{2}⁄_{5}, ^{2}⁄_{3}. In estimating for *lightness*, we observe that the greater the weight, the less the merit, so that *Z*’s merit is to *L*’s as 5 to 1. Thus the marks for *lightness* are ^{1}⁄_{5}, ^{2}⁄_{3}, 1. And similarly, the marks for warmth are 3, 1, ¼. To get the total result, we must *multiply* *L*’s 3 marks together, and do the same for *M* and for *Z*. The final numbers are 1 × ^{1}⁄_{5} × 3, ^{2}⁄_{5} × ^{2}⁄_{3} × 1, ^{2}⁄_{3} × 1 × ¼; *i.e.* ^{3}⁄_{5}, ^{2}⁄_{3}, ^{1}⁄_{3}; *i.e.* multiplying throughout by 15 (which will not alter the proportion), 9, 10, 5; showing the order of merit to be *M*, *L*, *Z*.

Twenty-nine answers have been received, of which five are right, and twenty-four wrong. These hapless ones have all (with three exceptions) fallen into the error of *adding* the proportional numbers together, for each candidate, instead of *multiplying*. *Why* the latter is right, rather than the former, is fully proved in text-books, so I will not occupy space by stating it here: but it can be *illustrated* very easily by the case of length, breadth, and depth. Suppose *A* and *B* are rival diggers of rectangular tanks: the amount of work done is evidently measured by the number of *cubical feet* dug out. Let *A* dig a tank 10 feet long, 10 wide, 2 deep: let *B* dig one 6 feet long, 5 wide, 10 deep. The cubical contents are 200, 300; *i.e.* *B* is best digger in the ratio of 3 to 2. Now try marking for length, width, and depth, separately; giving a maximum mark of 10 to the best in each contest, and then *adding* the results!

Of the twenty-four malefactors, one gives no working, and so has no real claim to be named; but I break the rule for once, in deference to its success in Problem 1: he, she, or it, is Addlepate. The other twenty-three may be divided into five groups.

First and worst are, I take it, those who put the rightful winner *last*; arranging them as "Lolo, Zuzu, Mimi." The names of these desperate wrong-doers are Ayr, Bradshaw of the Future, Furze-bush and Pollux (who send a joint answer), Greystead, Guy, Old Hen, and Simple Susan. The latter was *once* best of all; the Old Hen has taken advantage of her simplicity, and beguiled her with the chaff which was the bane of her own chickenhood.

Secondly, I point the finger of scorn at those who have put the worst candidate at the top; arranging them as "Zuzu, Mimi, Lolo." They are Graecia, M. M., Old Cat, and R. E. X. "’Tis Greece, but——."

The third set have avoided both these enormities, and have even succeeded in putting the worst last, their answer being "Lolo, Mimi, Zuzu." Their names are Ayr (who also appears among the "quite too too"), Clifton C., F. B., Fifee, Grig, Janet, and Mrs. Sairey Gamp. F. B. has not fallen into the common error; she *multiplies* together the proportionate numbers she gets, but in getting them she goes wrong, by reckoning warmth as a *de*-merit. Possibly she is "Freshly Burnt," or comes "From Bombay." Janet and Mrs. Sairey Gamp have also avoided this error: the method they have adopted is shrouded in mystery—I scarcely feel competent to criticize it. Mrs. Gamp says "if Zuzu makes 4 while Lolo makes 3, Zuzu makes 6 while Lolo makes 5 (bad reasoning), while Mimi makes 2." From this she concludes "therefore Zuzu excels in speed by 1" (*i.e.* when compared with Lolo; but what about Mimi?). She then compares the 3 kinds of excellence, measured on this mystic scale. Janet takes the statement, that "Lolo makes 5 while Mimi makes 2," to prove that "Lolo makes 3 while Mimi makes 1 and Zuzu 4" (worse reasoning than Mrs. Gamp’s), and thence concludes that "Zuzu excels in speed by ^{1}⁄_{8}"! Janet should have been Adeline, "mystery of mysteries!"

The fourth set actually put Mimi at the top, arranging them as "Mimi, Zuzu, Lolo." They are Marquis and Co., Martreb, S. B. B. (first initial scarcely legible: *may* be meant for "J"), and Stanza.

The fifth set consist of An ancient Fish and Camel. These ill-assorted comrades, by dint of foot and fin, have scrambled into the right answer, but, as their method is wrong, of course it counts for nothing. Also An ancient Fish has very ancient and fishlike ideas as to *how* numbers represent merit: she says "Lolo gains 2½ on Mimi." Two and a half *what*? Fish, fish, art thou in thy duty?

Of the five winners I put Balbus and The elder Traveller slightly below the other three—Balbus for defective reasoning, the other for scanty working. Balbus gives two reasons for saying that *addition* of marks is *not* the right method, and then adds "it follows that the decision must be made by *multiplying* the marks together." This is hardly more logical than to say "This is not Spring: *therefore* it must be Autumn."

CLASS LIST.

I.

Dinah Mite.

E. B. D. L.

Joram.

II.

Balbus.

The Elder Traveller.

With regard to Knot V., I beg to express to Vis Inertiæ and to any others who, like her, understood the condition to be that *every* marked picture must have *three* marks, my sincere regret that the unfortunate phrase "*fill* the columns with oughts and crosses" should have caused them to waste so much time and trouble. I can only repeat that a *literal* interpretation of "fill" would seem to *me* to require that *every* picture in the gallery should be marked. Vis Inertiæ would have been in the First Class if she had sent in the solution she now offers.

*Problem.*—Given that one glass of lemonade, 3 sandwiches, and 7 biscuits, cost 1*s.* 2*d.*; and that one glass of lemonade, 4 sandwiches, and 10 biscuits, cost 1*s.* 5*d.*: find the cost of (1) a glass of lemonade, a sandwich, and a biscuit; and (2) 2 glasses of lemonade, 3 sandwiches, and 5 biscuits.

*Answer.*—(1) 8*d.*; (2) 1*s.* 7*d.*

*Solution.*—This is best treated algebraically. Let *x* = the cost (in pence) of a glass of lemonade, *y* of a sandwich, and *z* of a biscuit. Then we have *x* + 3*y* + 7*z* = 14, and *x* + 4*y* + 10*z* = 17. And we require the values of *x* + *y* + *z*, and of 2*x* + 3*y* + 5*z*. Now, from *two* equations only, we cannot find, *separately*, the values of *three* unknowns: certain *combinations* of them may, however, be found. Also we know that we can, by the help of the given equations, eliminate 2 of the 3 unknowns from the quantity whose value is required, which will then contain one only. If, then, the required value is ascertainable at all, it can only be by the 3rd unknown vanishing of itself: otherwise the problem is impossible.

Let us then eliminate lemonade and sandwiches, and reduce everything to biscuits—a state of things even more depressing than "if all the world were apple-pie"—by subtracting the 1st equation from the 2nd, which eliminates lemonade, and gives *y* + 3*z* = 3, or *y* = 3-3*z*; and then substituting this value of *y* in the 1st, which gives *x*-2*z* = 5, *i.e.* *x* = 5 + 2*z*. Now if we substitute these values of *x*, *y*, in the quantities whose values are required, the first becomes (5 + 2*z*) + (3-3*z*) + *z*, *i.e.* 8: and the second becomes 2(5 + 2*z*) + 3(3-3*z*) + 5*z*, *i.e.* 19. Hence the answers are (1) 8*d.*, (2) 1*s.* 7*d.*

The above is a *universal* method: that is, it is absolutely certain either to produce the answer, or to prove that no answer is possible. The question may also be solved by combining the quantities whose values are given, so as to form those whose values are required. This is merely a matter of ingenuity and good luck: and as it *may* fail, even when the thing is possible, and is of no use in proving it *im*possible, I cannot rank this method as equal in value with the other. Even when it succeeds, it may prove a very tedious process. Suppose the 26 competitors, who have sent in what I may call *accidental* solutions, had had a question to deal with where every number contained 8 or 10 digits! I suspect it would have been a case of "silvered is the raven hair" (see "Patience") before any solution would have been hit on by the most ingenious of them.

Forty-five answers have come in, of which 44 give, I am happy to say, some sort of *working*, and therefore deserve to be mentioned by name, and to have their virtues, or vices as the case may be, discussed. Thirteen have made assumptions to which they have no right, and so cannot figure in the Class-list, even though, in 10 of the cases, the answer is right. Of the remaining 28, no less than 26 have sent in *accidental* solutions, and therefore fall short of the highest honours.

I will now discuss individual cases, taking the worst first, as my custom is.

Froggy gives no working—at least this is all he gives: after stating the given equations, he says "therefore the difference, 1 sandwich + 3 biscuits, = 3*d.*": then follow the amounts of the unknown bills, with no further hint as to how he got them. Froggy has had a *very* narrow escape of not being named at all!

Of those who are wrong, Vis Inertiæ has sent in a piece of incorrect working. Peruse the horrid details, and shudder! She takes *x* (call it "*y*") as the cost of a sandwich, and concludes (rightly enough) that a biscuit will cost (3-*y*)/3. She then subtracts the second equation from the first, and deduces 3*y* + 7 × (3-*y*)/3-4*y* + 10 × (3-*y*)/3 = 3. By making two mistakes in this line, she brings out *y* = ^{2}⁄_{2}. Try it again, oh Vis Inertiæ! Away with Inertiæ: infuse a little more Vis: and you will bring out the correct (though uninteresting) result, 0 = 0! This will show you that it is hopeless to try to coax any one of these 3 unknowns to reveal its *separate* value. The other competitor, who is wrong throughout, is either J. M. C. or T. M. C.: but, whether he be a Juvenile Mis-Calculator or a True Mathematician Confused, he makes the answers 7*d.* and 1*s.* 5*d.* He assumes, with Too Much Confidence, that biscuits were ½*d.* each, and that Clara paid for 8, though she only ate 7!

We will now consider the 13 whose working is wrong, though the answer is right: and, not to measure their demerits too exactly, I will take them in alphabetical order. Anita finds (rightly) that "1 sandwich and 3 biscuits cost 3*d.*," and proceeds "therefore 1 sandwich = 1½*d.*, 3 biscuits = 1½*d.*, 1 lemonade = 6*d.*" Dinah Mite begins like Anita: and thence proves (rightly) that a biscuit costs less than a 1*d.*: whence she concludes (wrongly) that it *must* cost ½*d.* F. C. W. is so beautifully resigned to the certainty of a verdict of "guilty," that I have hardly the heart to utter the word, without adding a "recommended to mercy owing to extenuating circumstances." But really, you know, where *are* the extenuating circumstances? She begins by assuming that lemonade is 4*d.* a glass, and sandwiches 3*d.* each, (making with the 2 given equations, *four* conditions to be fulfilled by *three* miserable unknowns!). And, having (naturally) developed this into a contradiction, she then tries 5*d.* and 2*d.* with a similar result. (N.B. *This* process might have been carried on through the whole of the Tertiary Period, without gratifying one single Megatherium.) She then, by a "happy thought," tries half-penny biscuits, and so obtains a consistent result. This may be a good solution, viewing the problem as a conundrum: but it is *not* scientific. Janet identifies sandwiches with biscuits! "One sandwich + 3 biscuits" she makes equal to "4." Four *what*? Mayfair makes the astounding assertion that the equation, *s* + 3*b* = 3, "is evidently only satisfied by *s* = ^{2}⁄_{2}, *b* = ½"! Old Cat believes that the assumption that a sandwich costs 1½*d.* is "the only way to avoid unmanageable fractions." But *why* avoid them? Is there not a certain glow of triumph in taming such a fraction? "Ladies and gentlemen, the fraction now before you is one that for years defied all efforts of a refining nature: it was, in a word, hopelessly vulgar. Treating it as a circulating decimal (the treadmill of fractions) only made matters worse. As a last resource, I reduced it to its lowest terms, and extracted its square root!" Joking apart, let me thank Old Cat for some very kind words of sympathy, in reference to a correspondent (whose name I am happy to say I have now forgotten) who had found fault with me as a discourteous critic. O. V. L. is beyond my comprehension. He takes the given equations as (1) and (2): thence, by the process [(2)-(1)] deduces (rightly) equation (3) viz. *s* + 3*b* = 3: and thence again, by the process [x3] (a hopeless mystery), deduces 3*s* + 4*b* = 4. I have nothing to say about it: I give it up. Sea-Breeze says "it is immaterial to the answer" (why?) "in what proportion 3*d.* is divided between the sandwich and the 3 biscuits": so she assumes *s* = l½*d.*, *b* = ½*d.* Stanza is one of a very irregular metre. At first she (like Janet) identifies sandwiches with biscuits. She then tries two assumptions (*s* = 1, *b* = ^{2}⁄_{3}, and *s* = ½ *b* = ^{2}⁄_{6}), and (naturally) ends in contradictions. Then she returns to the first assumption, and finds the 3 unknowns separately: *quod est absurdum*. Stiletto identifies sandwiches and biscuits, as "articles." Is the word ever used by confectioners? I fancied "What is the next article, Ma’am?" was limited to linendrapers. Two Sisters first assume that biscuits are 4 a penny, and then that they are 2 a penny, adding that "the answer will of course be the same in both cases." It is a dreamy remark, making one feel something like Macbeth grasping at the spectral dagger. "Is this a statement that I see before me?" If you were to say "we both walked the same way this morning," and *I* were to say "*one* of you walked the same way, but the other didn’t," which of the three would be the most hopelessly confused? Turtle Pyate (what *is* a Turtle Pyate, please?) and Old Crow, who send a joint answer, and Y. Y., adopt the same method. Y. Y. gets the equation *s* + 3*b* = 3: and then says "this sum must be apportioned in one of the three following ways." It *may* be, I grant you: but Y. Y. do you say "must"? I fear it is *possible* for Y. Y. to be *two* Y’s. The other two conspirators are less positive: they say it "can" be so divided: but they add "either of the three prices being right"! This is bad grammar and bad arithmetic at once, oh mysterious birds!

Of those who win honours, The Shetland Snark must have the 3rd class all to himself. He has only answered half the question, viz. the amount of Clara’s luncheon: the two little old ladies he pitilessly leaves in the midst of their "difficulty." I beg to assure him (with thanks for his friendly remarks) that entrance-fees and subscriptions are things unknown in that most economical of clubs, "The Knot-Untiers."

The authors of the 26 "accidental" solutions differ only in the number of steps they have taken between the *data* and the answers. In order to do them full justice I have arranged the 2nd class in sections, according to the number of steps. The two Kings are fearfully deliberate! I suppose walking quick, or taking short cuts, is inconsistent with kingly dignity: but really, in reading Theseus’ solution, one almost fancied he was "marking time," and making no advance at all! The other King will, I hope, pardon me for having altered "Coal" into "Cole." King Coilus, or Coil, seems to have reigned soon after Arthur’s time. Henry of Huntingdon identifies him with the King Coël who first built walls round Colchester, which was named after him. In the Chronicle of Robert of Gloucester we read:—

"Aftur Kyng Aruirag, of wam we habbeth y told,

Marius ys sone was kyng, quoynte mon & bold.

And ys sone was aftur hym,*Coil* was ys name,

Bothe it were quoynte men, & of noble fame."

Marius ys sone was kyng, quoynte mon & bold.

And ys sone was aftur hym,

Bothe it were quoynte men, & of noble fame."

Balbus lays it down as a general principle that "in order to ascertain the cost of any one luncheon, it must come to the same amount upon two different assumptions." (*Query.* Should not "it" be "we"? Otherwise the *luncheon* is represented as wishing to ascertain its own cost!) He then makes two assumptions—one, that sandwiches cost nothing; the other, that biscuits cost nothing, (either arrangement would lead to the shop being inconveniently crowded!)—and brings out the unknown luncheons as 8*d.* and 19*d.*, on each assumption. He then concludes that this agreement of results "shows that the answers are correct." Now I propose to disprove his general law by simply giving *one* instance of its failing. One instance is quite enough. In logical language, in order to disprove a "universal affirmative," it is enough to prove its contradictory, which is a "particular negative." (I must pause for a digression on Logic, and especially on Ladies’ Logic. The universal affirmative "everybody says he’s a duck" is crushed instantly by proving the particular negative "Peter says he’s a goose," which is equivalent to "Peter does *not* say he’s a duck." And the universal negative "nobody calls on her" is well met by the particular affirmative "*I* called yesterday." In short, either of two contradictories disproves the other: and the moral is that, since a particular proposition is much more easily proved than a universal one, it is the wisest course, in arguing with a Lady, to limit one’s *own* assertions to "particulars," and leave *her* to prove the "universal" contradictory, if she can. You will thus generally secure a *logical* victory: a *practical* victory is not to be hoped for, since she can always fall back upon the crushing remark "*that* has nothing to do with it!"—a move for which Man has not yet discovered any satisfactory answer. Now let us return to Balbus.) Here is my "particular negative," on which to test his rule. Suppose the two recorded luncheons to have been "2 buns, one queen-cake, 2 sausage-rolls, and a bottle of Zoëdone: total, one-and-ninepence," and "one bun, 2 queen-cakes, a sausage-roll, and a bottle of Zoëdone: total, one-and-fourpence." And suppose Clara’s unknown luncheon to have been "3 buns, one queen-cake, one sausage-roll, and 2 bottles of Zoëdone:" while the two little sisters had been indulging in "8 buns, 4 queen-cakes, 2 sausage-rolls, and 6 bottles of Zoëdone." (Poor souls, how thirsty they must have been!) If Balbus will kindly try this by his principle of "two assumptions," first assuming that a bun is 1*d.* and a queen-cake 2*d.*, and then that a bun is 3*d.* and a queen-cake 3*d.*, he will bring out the other two luncheons, on each assumption, as "one-and-nine-pence" and "four-and-ten-pence" respectively, which harmony of results, he will say, "shows that the answers are correct." And yet, as a matter of fact, the buns were 2*d.* each, the queen-cakes 3*d.*, the sausage-rolls 6*d.*, and the Zoëdone 2*d.* a bottle: so that Clara’s third luncheon had cost one-and-sevenpence, and her thirsty friends had spent four-and-fourpence!

Another remark of Balbus I will quote and discuss: for I think that it also may yield a moral for some of my readers. He says "it is the same thing in substance whether in solving this problem we use words and call it Arithmetic, or use letters and signs and call it Algebra." Now this does not appear to me a correct description of the two methods: the Arithmetical method is that of "synthesis" only; it goes from one known fact to another, till it reaches its goal: whereas the Algebraical method is that of "analysis": it begins with the goal, symbolically represented, and so goes backwards, dragging its veiled victim with it, till it has reached the full daylight of known facts, in which it can tear off the veil and say "I know you!"

Take an illustration. Your house has been broken into and robbed, and you appeal to the policeman who was on duty that night. "Well, Mum, I did see a chap getting out over your garden-wall: but I was a good bit off, so I didn’t chase him, like. I just cut down the short way to the Chequers, and who should I meet but Bill Sykes, coming full split round the corner. So I just ups and says ’My lad, you’re wanted.’ That’s all I says. And he says ’I’ll go along quiet, Bobby,’ he says, ’without the darbies,’ he says." There’s your *Arithmetical* policeman. Now try the other method. "I seed somebody a running, but he was well gone or ever *I* got nigh the place. So I just took a look round in the garden. And I noticed the foot-marks, where the chap had come right across your flower-beds. They was good big foot-marks sure-ly. And I noticed as the left foot went down at the heel, ever so much deeper than the other. And I says to myself ’The chap’s been a big hulking chap: and he goes lame on his left foot.’ And I rubs my hand on the wall where he got over, and there was soot on it, and no mistake. So I says to myself ’Now where can I light on a big man, in the chimbley-sweep line, what’s lame of one foot?’ And I flashes up permiscuous: and I says ’It’s Bill Sykes!’ says I." There is your *Algebraical* policeman—a higher intellectual type, to my thinking, than the other.

Little Jack’s solution calls for a word of praise, as he has written out what really is an algebraical proof *in words*, without representing any of his facts as equations. If it is all his own, he will make a good algebraist in the time to come. I beg to thank Simple Susan for some kind words of sympathy, to the same effect as those received from Old Cat.

Hecla and Martreb are the only two who have used a method *certain* either to produce the answer, or else to prove it impossible: so they must share between them the highest honours.

CLASS LIST.

I.

Hecla.

Martreb.

II.

§ 1 (2 *steps*).

Adelaide.

Clifton C....

E. K. C.

Guy.

L’Inconnu.

Little Jack.

Nil desperandum.

Simple Susan.

Yellow-Hammer.

Woolly One.

§ 2 (3 *steps*).

A. A.

A Christmas Carol.

Afternoon Tea.

An appreciative Schoolma’am.

Baby.

Balbus.

Bog-Oak.

The Red Queen.

Wall-flower.

§ 3 (4 *steps*).

Hawthorn.

Joram.

S. S. G.

§ 4 (5 *steps*).

A Stepney Coach.

§ 5 (6 *steps*).

Bay Laurel.

Bradshaw of the Future.

§ 6 (9 *steps*).

Old King Cole.

§ 7 (14 *steps*).

Theseus.