"Aftur Kyng Aruirag, of wam we habbeth y told,Marius ys sone was kyng, quoynte mon & bold.And ys sone was aftur hym,Coilwas ys name,Bothe it were quoynte men, & of noble fame."
"Aftur Kyng Aruirag, of wam we habbeth y told,Marius ys sone was kyng, quoynte mon & bold.And ys sone was aftur hym,Coilwas ys name,Bothe it were quoynte men, & of noble fame."
Balbuslays 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 theluncheonis 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 unknownluncheons as 8d.and 19d., 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 givingoneinstance 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 doesnotsay he's a duck." And the universal negative "nobody calls on her" is well met by the particular affirmative "Icalled 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'sownassertions to "particulars," and leaveherto prove the "universal" contradictory, if she can. You will thus generally secure alogicalvictory: apracticalvictory is not to be hoped for, since she can always fall back upon the crushing remark "thathas nothing to do with it!"—a move for which Man has not yet discovered any satisfactory answer. Now let us return toBalbus.) Here is my "particular negative," on which to test his rule. Suppose the tworecorded 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!) IfBalbuswill kindly try this by his principle of "two assumptions," first assuming that a bun is 1d.and a queen-cake 2d., and then that a bun is 3d.and a queen-cake 3d., 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 2d.each, the queen-cakes 3d., the sausage-rolls 6d., and the Zoëdone 2d.a bottle: so that Clara's third luncheon had cost one-and-sevenpence, and her thirsty friends had spentfour-and-fourpence!
Another remark ofBalbusI 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 yourArithmeticalpoliceman. Now try the other method. "I seed somebody a running, but he was well gone or everIgot 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 yourAlgebraicalpoliceman—a higher intellectual type, to my thinking, than the other.
Little Jack'ssolution calls for a word of praise, as he has written out what really is an algebraical proofin 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 thankSimple Susanfor some kind words of sympathy, to the same effect as those received fromOld Cat.
HeclaandMartrebare the only two who have used a methodcertaineither 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 (2steps).
Adelaide.Clifton C....E. K. C.Guy.L'Inconnu.Little Jack.Nil desperandum.Simple Susan.Yellow-Hammer.Woolly One.
§ 2 (3steps).
A. A.A Christmas Carol.Afternoon Tea.An appreciative Schoolma'am.Baby.Balbus.Bog-Oak.The Red Queen.Wall-flower.
§ 3 (4steps).
Hawthorn.Joram.S. S. G.
§ 4 (5steps).
A Stepney Coach.
§ 5 (6steps).
Bay Laurel.Bradshaw of the Future.
§ 6 (9steps).
Old King Cole.
§ 7 (14steps).
Theseus.
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.Trojanushowever 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 merejeu 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," saidhe, "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,AandB, who are to compete in French, German, and Italian:
(a) Let it be laid down that the result is to dependon theirrelativeknowledge 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 multiplyA's 3 marks together, and do the same forB. Note that, ifAgets a single "0," his final mark is "0," even if he gets full marks for 2 papers whileBgets only one mark for each paper. This of course would be very unfair onA, though a correct solution under the given conditions.
(b) The result is to depend, as before, onrelativeknowledge; 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 multiplierstwiceto produce the same result as in (a):"e.g.if the French Ratio were2⁄10, and the others2⁄9,1⁄9so that the ultimate ratio, by method (a), would be2⁄45, I should multiply instead by2⁄3,1⁄3, giving the result,1⁄3which is nearer to2⁄10than if he had used method (a).
(c) The result is to depend onactualamount of knowledge of the 3 subjects collectively. Here we haveto 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 addA's 3 marks together, and do the same forB.
(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 addA'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 ofscarves 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 toBalbus, 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 toFifee, who says "I think a little more consideration will show you that our 'error ofaddingthe proportional numbers together for each candidate instead ofmultiplying' is no error at all." Why, even if additionhadbeen 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 tothe 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 ofbeing correct." I trust that the knot-untiers and I are not on such terms as those!
I beg to thankG. 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.Balbussays "it certainly cannot be solved mathematically, nor do I see how to solve it by any verbal quibble."Nolens Volensmakes 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 onfoot 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." Hencea+x= 5x;i.e.4x=a, andx=a/4. This distance would be traversed by an omnibus in15⁄4minutes, 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 Miterightly 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 aheadof 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 Volenstries 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 1-12th. Theprincipleis 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.Heclasays 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 toHecla, 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.Magpiesays 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 Coletakes the same view asHecla.TympanumandVindexassume 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 asHecla.
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.
Tympanumsays that the statement about the stick "is merely a blind, to which the old answer may well be applied,solvitur ambulando, or rathermergendo." I trustTympanumwill not test this in his own person, by taking the place of the man in Balbus' Essay! He would infallibly be drowned.
Old King Colerightly points out that the series, 2, 1, &c., is a decreasing Geometrical Progression: whileVindexrightly 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., ifxbe the width,x(x+ ½) = 3,630. Solving this Quadratic, we findx= 60. Hence the dimensions are 60, 60½.
Twelve answers have been received—seven right and five wrong.
C. G. L.,Nabob,Old Crow, andTympanumassume 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'sis shorter, and so (if possible) worth rather less. He says the answer "is at once seen to be 60 × 60½"!Nabob'scalculation is short, but "as rich as a Nabob" in error. He says that the square root of 3,630, multiplied by 2, equals thelength plus the breadth. That is 60.25 × 2 = 120½. His first assertion is only true of asquaregarden. His second is irrelevant, since 60.25 isnotthe square-root of 3,630! Nay, Bob, this willnotdo!Tympanumsays 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.Tympanumproceeds 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.
Heclaindulges, again and again, in that most fatal of all habits in computation—the makingtwomistakes which cancel each other. She takesxas the width of the garden, in yards, andx+ ½ as its length, and makes her first "coil" the sum ofx½,x½,x-1,x-1,i.e.4x-3: but the fourth term should bex-1½, so that her first coil is ½ a yard too long. Her second coil is the sum ofx-2½,x-2½,x-3,x-3: here the first term should bex-2 and the lastx-3½: these twomistakes 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 theendof 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, andTaffymake the same assumption as C. G. L. and Co. They then solve by a Quadratic.Magpiealso tries it by Arithmetical Progression, but fails to notice that the first and last "coils" have special values.
Alumnus Etonæattempts 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 Colesolves it by an Arithmetical Progression. It is right, but too lengthy to be worth as much as a Quadratic.
Vindexproves 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.
§ 1.The Chelsea Pensioners.
Problem.—If 70 per cent. have lost an eye, 75 per cent. an ear, 80 per cent. an arm, 85 per cent. a leg: what percentage,at least, must have lost all four?
Answer.—Ten.
Solution.—(I adopt that ofPolar Star, as being better than my own). Adding the wounds together, we get 70 + 75 + 80 + 85 = 310, among 100 men; which gives 3 to each, and 4 to 10 men. Therefore the least percentage is 10.
Nineteen answers have been received. One is "5," but, as no working is given with it, it must, in accordance with the rule, remain "a deed without a name."Janetmakes it "35 and2⁄10ths." I am sorry she has misunderstood the question, and has supposed that those who had lost an ear were 75 per cent.of those who had lost an eye; and so on. Of course, on this supposition, the percentages must all be multiplied together. This she hasdone correctly, but I can give her no honours, as I do not think the question will fairly bear her interpretation,Three Score and Tenmakes it "19 and2⁄8ths." Her solution has given me—I will not say "many anxious days and sleepless nights," for I wish to be strictly truthful, but—some trouble in making any sense at all of it. She makes the number of "pensioners wounded once" to be 310 ("per cent.," I suppose!): dividing by 4, she gets 77 and a half as "average percentage:" again dividing by 4, she gets 19 and2⁄8ths as "percentage wounded four times." Does she suppose wounds of different kinds to "absorb" each other, so to speak? Then, no doubt, thedataare equivalent to 77 pensioners with one wound each, and a half-pensioner with a half-wound. And does she then suppose these concentrated wounds to betransferable, so that2⁄4ths of these unfortunates can obtain perfect health by handing over their wounds to the remaining1⁄4th? Granting these suppositions, her answer is right; or rather,ifthe question had been "A road is covered with one inch of gravel, along 77 and a half per cent. of it. How much of it could be covered 4 inches deep with the same material?" her answerwouldhave been right. But alas, thatwasn'tthe question!Deltamakes some most amazing assumptions: "let every one who has not lost an eye have lost an ear," "let every one who has not lost both eyes and ears have lost an arm."Her ideas of a battle-field are grim indeed. Fancy a warrior who would continue fighting after losing both eyes, both ears, and both arms! This is a case which she (or "it?") evidently considerspossible.
Next come eight writers who have made the unwarrantable assumption that, because 70 per cent. have lost an eye,therefore30 per cent. havenotlost one, so that they havebotheyes. This is illogical. If you give me a bag containing 100 sovereigns, and if in an hour I come to you (my facenotbeaming with gratitude nearly so much as when I received the bag) to say "I am sorry to tell you that 70 of these sovereigns are bad," do I thereby guarantee the other 30 to be good? Perhaps I have not tested them yet. The sides of this illogical octagon are as follows, in alphabetical order:—Algernon Bray,Dinah Mite, G. S. C.,Jane E., J. D. W.,Magpie(who makes the delightful remark "therefore 90 per cent. have two of something," recalling to one's memory that fortunate monarch, with whom Xerxes was so much pleased that "he gave him ten of everything!"), S. S. G., andTokio.
Bradshaw of the Futureand T. R. do the question in a piecemeal fashion—on the principle that the 70 per cent. and the 75 per cent., though commenced at opposite ends of the 100, must overlap byat least45 per cent.; and so on. This is quite correct working, but not, I think, quite the best way of doing it.
The other five competitors will, I hope, feel themselves sufficiently glorified by being placed in the first class, without my composing a Triumphal Ode for each!
CLASS LIST.
I.
Old Cat.Old Hen.Polar Star.Simple Susan.White Sugar.
II.
Bradshaw of the Future.T. R.
III.
Algernon Bray.Dinah Mite.G. S. C.Jane E.J. D. W.Magpie.S. S. G.Tokio.
§ 2.Change of Day.
I must postpone,sine die, the geographical problem—partly because I have not yet received the statistics I am hoping for, and partly because I am myself so entirely puzzled by it; and when an examiner is himself dimly hovering between a second class and a third how is he to decide the position of others?
§ 3.The Sons' Ages.
Problem.—"At first, two of the ages are together equal to the third. A few years afterwards, two of them are together double of the third. When the number of years since the first occasion is two-thirds of the sum of the ages on that occasion, one age is 21. What are the other two?
Answer.—"15 and 18."
Solution.—Let the ages at first bex,y, (x+y). Now, ifa+b= 2c, then (a-n) + (b-n) = 2(c-n), whatever be the value ofn. Hence the second relationship, ifevertrue, wasalwaystrue. Hence it was true at first. But it cannot be true thatxandyare together double of (x+y). Hence it must be true of (x+y), together withxory; and it does not matter which we take. We assume, then, (x+y) +x= 2y;i.e.y= 2x. Hence the three ages were, at first,x, 2x, 3x; and the number of years, since that time is two-thirds of 6x,i.e.is 4x. Hence the present ages are 5x, 6x, 7x. The ages are clearlyintegers, since this is only "the year when one of my sons comes of age." Hence 7x= 21,x= 3, and the other ages are 15, 18.
Eighteen answers have been received. One of the writers merely asserts that the first occasion was 12 years ago, that the ages were then 9, 6, and 3; and that on the second occasion they were 14, 11, and 8! As a Roman father, Ioughtto withhold the name of the rash writer; but respect for age makes me break the rule: it isThree Score and Ten.Jane E.also asserts that the ages at first were 9, 6, 3: then she calculates the present ages, leaving thesecondoccasion unnoticed.Old Henis nearly as bad; she "tried various numbers till I found one that fittedallthe conditions"; but merely scratching up the earth, and pecking about, isnotthe way to solve a problem, oh venerable bird! And close afterOld Henprowls, with hungry eyes,Old Cat, who calmly assumes, to begin with, that the son who comes of age is theeldest. Eat your bird, Puss, for you will get nothing from me!
There are yet two zeroes to dispose of.Minervaassumes that, oneveryoccasion, a son comes of age; and that it is only such a son who is "tipped with gold." Is it wise thus to interpret "now, my boys, calculate your ages, and you shall have the money"?Bradshaw of the Futuresays "let" the ages at first be 9, 6, 3, then assumes that the second occasion was 6 years afterwards, and on these baseless assumptions brings out the rightanswers. Guidefuturetravellers, an thou wilt: thou art no Bradshaw forthisAge!
Of those who win honours, the merely "honourable" are two.Dinah Miteascertains (rightly) the relationship between the three ages at first, but thenassumesone of them to be "6," thus making the rest of her solution tentative. M. F. C. does the algebra all right up to the conclusion that the present ages are 5z, 6z, and 7z; it then assumes, without giving any reason, that 7z= 21.
Of the more honourable,Deltaattempts a novelty—to discoverwhichson comes of age by elimination: it assumes, successively, that it is the middle one, and that it is the youngest; and in each case itapparentlybrings out an absurdity. Still, as the proof contains the following bit of algebra, "63 = 7x+ 4y; ∴ 21 =x+ 4 sevenths ofy," I trust it will admit that its proof is notquiteconclusive. The rest of its work is good.Magpiebetrays the deplorable tendency of her tribe—to appropriate any stray conclusion she comes across, without having anystrictlogical right to it. AssumingA,B,C, as the ages at first, andDas the number of the years that have elapsed since then, she finds (rightly) the 3 equations, 2A=B,C=B+A,D= 2B. She then says "supposing thatA= 1, thenB= 2,C= 3, andD= 4. Therefore forA,B,C,D, four numbers are wanted which shall be toeach other as 1:2:3:4." It is in the "therefore" that I detect the unconscientiousness of this bird. The conclusionistrue, but this is only because the equations are "homogeneous" (i.e.having one "unknown" in each term), a fact which I strongly suspect had not been grasped—I beg pardon, clawed—by her. Were I to lay this little pitfall, "A+ 1 =B,B+ 1 =C; supposingA= 1, thenB= 2 andC= 3.ThereforeforA,B,C, three numbers are wanted which shall be to one another as 1:2:3," would you not flutter down into it, ohMagpie, as amiably as a Dove?Simple Susanis anything but simple tome. After ascertaining that the 3 ages at first are as 3:2:1, she says "then, as two-thirds of their sum, added to one of them, = 21, the sum cannot exceed 30, and consequently the highest cannot exceed 15." I suppose her (mental) argument is something like this:—"two-thirds of sum, + one age, = 21; ∴ sum, + 3 halves of one age, = 31 and a half. But 3 halves of one age cannot be less than 1 and-a-half (here I perceive thatSimple Susanwould on no account present a guinea to a new-born baby!) hence the sum cannot exceed 30." This is ingenious, but her proof, after that, is (as she candidly admits) "clumsy and roundabout." She finds that there are 5 possible sets of ages, and eliminates four of them. Suppose that, instead of 5, there had been 5 million possible sets? WouldSimple Susanhavecourageously ordered in the necessary gallon of ink and ream of paper?
The solution sent in by C. R. is, like that ofSimple Susan, partly tentative, and so does not rise higher than being Clumsily Right.
Among those who have earned the highest honours,Algernon Braysolves the problem quite correctly, but adds that there is nothing to exclude the supposition that all the ages werefractional. This would make the number of answers infinite. Let me meekly protest that Ineverintended my readers to devote the rest of their lives to writing out answers!E. M. Rixpoints out that, if fractional ages be admissible, any one of the three sons might be the one "come of age"; but she rightly rejects this supposition on the ground that it would make the problem indeterminate.White Sugaris the only one who has detected an oversight of mine: I had forgotten the possibility (which of course ought to be allowed for) that the son, who came of age thatyear, need not have done so by thatday, so that hemightbe only 20. This gives a second solution, viz., 20, 24, 28. Well said, pure Crystal! Verily, thy "fair discourse hath been as sugar"!
CLASS LIST.
I.
Algernon Bray.An Old Fogey.E. M. Rix.G. S. C.S. S. G.Tokio.T. R.White Sugar.
II.
C. R.Delta.Magpie.Simple Susan.
III.
Dinah Mite.M. F. C.
I have received more than one remonstrance on my assertion, in the Chelsea Pensioners' problem, that it was illogical to assume, from thedatum"70 p. c. have lost an eye," that 30 p. c. havenot.Algernon Braystates, as a parallel case, "suppose Tommy's father gives him 4 apples, and he eats one of them, how many has he left?" and says "I think we are justified in answering, 3." I think so too. There is no "must" here, and thedataare evidently meant to fix the answerexactly: but, if the question were set me "how manymusthe have left?", I should understand thedatato be that his father gave him 4at least, butmayhave given him more.
I take this opportunity of thanking those who have sent, along with their answers to the Tenth Knot, regrets that there are no more Knots to come, or petitions that I should recall my resolution to bring them to an end. I am most grateful for their kind words; but I think it wisest to end what, at best, was but a lame attempt. "The stretched metre of an antique song" is beyond my compass; and my puppets were neither distinctlyinmy life (like those I now address), nor yet (like Alice and the Mock Turtle) distinctlyoutof it. Yet let me at least fancy, as I lay down the pen, that I carry with me into my silent life, dear reader, a farewell smile from your unseen face, and a kindly farewell pressure from your unfelt hand! And so, good night! Parting is such sweet sorrow, that I shall say "good night!" till it be morrow.
THE END
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