Chapter 5

The seventeen whose solutions are practically valueless are ARDMORE, A READY RECKONER, ARTHUR, BOG-LARK, BOG-OAK, BRIDGET, FIRST ATTEMPT, J. L. C., M. E. T., ROSE, ROWENA, SEA-BREEZE, SYLVIA, THISTLEDOWN, THREE-FIFTHS ASLEEP, VENDREDI, and WINIFRED. BOG-LARK tries it by a sort of "rule of false," assuming experimentally that Nos. 1, 2, weigh 6 lbs. each, and having thus produced 17-1/2, instead of 16, as the weight of 1, 3, and 5, she removes "the superfluous pound and a half," but does not explain how she knows from which to take it. THREE-FIFTHS ASLEEP says that (when in that peculiar state) "it seemed perfectly clear" to her that, "3 out of the 5 sacks being weighed twice over, 3/5 of 45 = 27, must be the total weight of the 5 sacks." As to which I can only say, with the Captain, "it beats me entirely!" WINIFRED, on the plea that "one must have a starting-point," assumes (what I fear is a mere guess) that No. 1 weighed 5-1/2 lbs. The rest all do it, wholly or partly, by guess-work.

The problem is of course (as any Algebraist sees at once) a case of "simultaneous simple equations." It is, however, easily soluble by Arithmetic only; and, when this is the case, I hold that it is bad workmanship to use the more complex method. I have not, this time, given more credit to arithmetical solutions; but in future problems I shall (other things being equal) give the highest marks to those who use the simplest machinery. I have put into Class I. those whose answers seemed specially short and neat, and into Class III. those that seemed specially long or clumsy. Of this last set, A. C. M., FURZE-BUSH, JAMES, PARTRIDGE, R. W., and WAITING FOR THE TRAIN, have sent long wandering solutions, the substitutions having no definite method, but seeming to have been made to see what would come of it. CHILPOME and DUBLIN BOY omit some of the working. ARVON MARLBOROUGH BOY only finds the weight of _one_ sack.

CLASS LIST

I.

B. E. D. C. H. CONSTANCE JOHNSON. GREYSTEAD. GUY. HOOPOE. J. F. A. M. A. H. NUMBER FIVE. PEDRO. R. E. X. SEVEN OLD MEN. VIS INERTIÆ. WILLY B. YAHOO.

II.

AMERICAN SUBSCRIBER. AN APPRECIATIVE SCHOOLMA'AM. AYR. BRADSHAW OF THE FUTURE. CHEAM. C. M. G. DINAH MITE. DUCKWING. E. C. M. E. N. Lowry. ERA. EUROCLYDON. F. H. W. FIFEE. G. E. B. HARLEQUIN. HAWTHORN. HOUGH GREEN. J. A. B. JACK TAR. J. B. B. KGOVJNI. LAND LUBBER. L. D. MAGPIE. MARY. MHRUXI. MINNIE. MONEY-SPINNER. NAIRAM. OLD CAT. POLICHINELLE. SIMPLE SUSAN. S. S. G. THISBE. VERENA. WAMBA. WOLFE. WYKEHAMICUS. Y. M. A. H.

III.

A. C. M. ARVON MARLBOROUGH BOY. CHILPOME. DUBLIN BOY. FURZE-BUSH. JAMES. PARTRIDGE. R. W. WAITING FOR THE TRAIN.

ANSWERS TO KNOT V.

_Problem._--To mark pictures, giving 3 x's to 2 or 3, 2 to 4 or 5, and 1 to 9 or 10; also giving 3 o's to 1 or 2, 2 to 3 or 4 and 1 to 8 or 9; so as to mark the smallest possible number of pictures, and to give them the largest possible number of marks.

_Answer._--10 pictures; 29 marks; arranged thus:--

x x x x x x x x x o x x x x x o o o o x x o o o o o o o o

_Solution._--By giving all the x's possible, putting into brackets the optional ones, we get 10 pictures marked thus:--

x x x x x x x x x (x) x x x x (x) x x (x)

By then assigning o's in the same way, beginning at the other end, we get 9 pictures marked thus:--

(o) o (o) o o o (o) o o o o o o o o

All we have now to do is to run these two wedges as close together as they will go, so as to get the minimum number of pictures----erasing optional marks where by so doing we can run them closer, but otherwise letting them stand. There are 10 necessary marks in the 1st row, and in the 3rd; but only 7 in the 2nd. Hence we erase all optional marks in the 1st and 3rd rows, but let them stand in the 2nd.

* * * * *

Twenty-two answers have been received. Of these 11 give no working; so, in accordance with what I announced in my last review of answers, I leave them unnamed, merely mentioning that 5 are right and 6 wrong.

Of the eleven answers with which some working is supplied, 3 are wrong. C. H. begins with the rash assertion that under the given conditions "the sum is impossible. For," he or she adds (these initialed correspondents are dismally vague beings to deal with: perhaps "it" would be a better pronoun), "10 is the least possible number of pictures" (granted): "therefore we must either give 2 x's to 6, or 2 o's to 5." Why "must," oh alphabetical phantom? It is nowhere ordained that every picture "must" have 3 marks! FIFEE sends a folio page of solution, which deserved a better fate: she offers 3 answers, in each of which 10 pictures are marked, with 30 marks; in one she gives 2 x's to 6 pictures; in another to 7; in the 3rd she gives 2 o's to 5; thus in every case ignoring the conditions. (I pause to remark that the condition "2 x's to 4 or 5 pictures" can only mean "_either_ to 4 _or else_ to 5": if, as one competitor holds, it might mean _any_ number not less than 4, the words "_or_ 5" would be superfluous.) I. E. A. (I am happy to say that none of these bloodless phantoms appear this time in the class-list. Is it IDEA with the "D" left out?) gives 2 x's to 6 pictures. She then takes me to task for using the word "ought" instead of "nought." No doubt, to one who thus rebels against the rules laid down for her guidance, the word must be distasteful. But does not I. E. A. remember the parallel case of "adder"? That creature was originally "a nadder": then the two words took to bandying the poor "n" backwards and forwards like a shuttlecock, the final state of the game being "an adder." May not "a nought" have similarly become "an ought"? Anyhow, "oughts and crosses" is a very old game. I don't think I ever heard it called "noughts and crosses."

In the following Class-list, I hope the solitary occupant of III. will sheathe her claws when she hears how narrow an escape she has had of not being named at all. Her account of the process by which she got the answer is so meagre that, like the nursery tale of "Jack-a-Minory" (I trust I. E. A. will be merciful to the spelling), it is scarcely to be distinguished from "zero."

CLASS LIST.

I.

GUY. OLD CAT. SEA-BREEZE.

II.

AYR. BRADSHAW OF THE FUTURE. F. LEE. H. VERNON.

III.

CAT.

ANSWERS TO KNOT VI.

_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, 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 × 1/5 × 3, 2/5 × 5/3 × 1, 4/3 × 1 × 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 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 2-1/2 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.

ANSWERS TO KNOT VII.

_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 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, = 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_ = 3/2. 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 1/2_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 = 1-1/2_d._, 3 biscuits = 1-1/2_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 1/2_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_ = 3/2, _b_ = 1/2"! OLD CAT believes that the assumption that a sandwich costs 1-1/2_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 [×33] (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_ = l-1/2_d._, _b_ = 1/2_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_ = 1/2 _b_ = 5/6), 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:--