A Tangled Tale, by Lewis Carroll

Answers to Knot 6

Problem 1. — A and B began the year with only £1000 apiece. They borrowed nought; they stole nought. On the next New Year’s Day they had £60,000 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 “zero” and steal “zero”, and uses both cyphers by putting them at the righthand end of the £1000, thus producing £100,000, 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 £1000 into a “9”, and adds the result to the original sum, thus getting £10,000: and in this, by means of the other “zero”, she turns the “1” into a “6” thus hitting the exact £60,000.

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 of 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, 2/3, 4/3. 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, 5/3, 1. And similarly, the marks for warmth are 3, 1, 1/4. 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 x 1/5 x 3, 2/5 x 5/3 x 1, 4/3 x 1 x 1/4; 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 textbooks, 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, 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., Fiffee, Grig, Janet, and Mrs. Sairey Gamp. F. B. has not fallen into the common error; she multiplies together the proportionate number she gets, but in getting them she goes wrong, by reckoning warmth as a demerit. Possibly she is “Freshly Burnt”, or comes “From Bombay” Janet and Mrs. Sairey Gamp have also avoided this error: the method thev have adopted is shrouded in mystery — I scarcely feel competent to criticise it. Mrs. Gamp says, “If Zuzu makes 4 while Lolo makes 3, Zuzu makes 6 while Lolo makes 5 [bad reasoningl, 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 consists 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 Inertiae 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 interpretstion of “fill” would seem to me to require that every picture in the gallery should be marked. Vis Inertiae would have been in the First Class if she had sent in the solution she now offers.

http://ebooks.adelaide.edu.au/c/carroll/lewis/tangled/answers6.html

Last updated Saturday, March 1, 2014 at 20:37