A Tangled Tale, Lewis Carroll [free ebook novel .txt] 📗

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 impossible, 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 13 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, = 3d.”: 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 3y+7×3−y3−4y+10×3−y3=3. By making two mistakes in this line, she brings out y=32. 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 7d. and 1s. 5d. He assumes, with Too Much Confidence, that biscuits were ½d. 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 3d.,” and proceeds “therefore 1 sandwich = 1½d., 3 biscuits = 1½d., 1 lemonade = 6d.” Dinah Mite begins like Anita: and thence proves (rightly) that a biscuit costs less than a 1d.: whence she concludes (wrongly) that it must cost ½d. 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 4d. a glass, and sandwiches 3d. 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 5d. and 2d. 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 halfpenny 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+3b=3, “is evidently only satisfied by s=32, b=12”! Old Cat believes that the assumption that a sandwich costs 1½d. 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+3b=3: and thence again, by the process [×3] (a hopeless mystery), deduces 3s+4b=4. I have nothing to say about it: I give it up. Sea-Breeze says “it is immaterial to the answer” (why?) “in