 Chapter 14 of A Tangled Tale Answers to not four There are five sacks of which numbers one, two, weigh twelve pounds, numbers two, three, thirteen and a half pounds, numbers three, four, eleven and a half pounds, numbers four, five, eight pounds, numbers one, three, five, sixteen pounds, required the weight of each sack. Answer, five and a half, six and a half, seven, four and a half, three and a half. The sum of all the weighings, sixty-one pounds, includes sack number three, thrice, and each other, twice. Deducting twice the sum of the first and fourth weighings, we get twenty-one pounds for thrice number three, that is, seven pounds for number three. Hence the second and third weighings give six and a half pounds, four and a half pounds for numbers two, four, and hence again the first and fourth weighings give five and a half pounds, three and a half pounds for numbers one, five. Ninety-seven answers have been received. Of these, fifteen are beyond the reach of discussion, as they give no working. I can but enumerate their names, and I take this opportunity of saying that this is the last time I shall put on record the names of competitors, who give no sort of clue to the process by which their answers were obtained. In guessing a conundrum, or in catching a flea, we do not expect a breathless victor to give us afterwards, in cold blood, a history of the mental or muscular efforts by which he achieved success. But a mathematical calculation is another thing. The names of these muting glorious band are common sense, D-E-R, Douglas, E-L, Ellen, I-M-T, J-M-C, Joseph, not one, Lucy, Meek, M-F-C, Pyramus, Shah, Veritas. Of the eighty-two answers with which the working or some approach to it is supplied, one is wrong. Seventeen have given solutions which are, from one cause or another, practically valueless. The remaining sixty-four I shall try to arrange in a class list, according to the varying degrees of shortness and needness to which they seem to have attained. The solitary wrong answer is from Nell. To be thus alone in the crowd is a distinction. A painful one, no doubt, but still a distinction. I am sorry for you, my dear young lady, and I seem to hear your tearful exclamation when you read these lines. Ah, this is the Nell of all my hopes. Why or why did you assume that the fourth and fifth bags weighed four pounds each? And why did you not test your answers? However, please try again, and please don't change your nom de plume. Let us have Nell in the first class next time. The seventeen, whose solutions are practically valueless, are Artmore, Reddy Reckoner, Arthur, Boglark, Bog Oak, Bridget, first attempt. JLC, M-E-T, Rose, Rovina, Seabreeze, Sylvia, Thistaldown, three-fifths asleep, Vendredi and Winifred. Boglark tries it by a sort of rule of false, assuming experimentally that numbers one, two weigh six pounds each, and having thus produced seventeen and a half instead of sixteen as the weight of one, three and five, 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 seems perfectly clear to her that three out of the five sacks being weighed twice over, two-fifths of forty-five equals twenty-seven, must be the total weight of the five 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 number one weighed five and a half pounds. The rest I'll do it wholly or partly by guesswork. 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 one those whose answers seemed specially short and neat, and into class three those that seemed specially long or clumsy. Of this last set, ACM, Firstbush, James, Partridge, RW, and Waiting for the Train have sent long-wondering solutions. The substitutions having no definite method, but seeming to have been made to see what would come of it. Chill-Poam and Dublin Boy omit some of the working. Arvin Marlborough Boy only finds the weight of one sack. Class List First B.E.D. C.H. Constance Johnson Graystead Guy Hoopoe J.F.A. M.A.H. Number Five Pedro R.E.X. Seven Old Men Vis Inertier Willi B Yahoo Second American Subscriber An Appreciative School Ma'am Iyer Bradshaw of the Future Chim CMG Dynamite Duck Wing E.C.M Ian Lowry Ira Euro Clyden F.H.W. Fifi G.E.B Hallikin Hawthorne Hugh Green J.A.B Jack Tarr J.B.B. Kugovdni Landlubber L.D. Magpie Mary Ruxi Mini Mane Spinner Nairam Old Cat Polychannel Simple Susan SSG Thisp Verena Wamba Wolff Y.Chemicus Y.M.A.H. Third ACM Arvon Marlboro Boy Chilpom Dublin Boy Fersbush James Partridge R.W. Waiting for the train End of Chapter 14 Chapter 15 of A Tangled Tale This LibriVox recording is in the public domain. Recording by Avae in October 2009. A Tangled Tale by Lewis Carroll Chapter 15 Answers to not five Problem To mark pictures Giving three crosses to two or three Two to four or five And one to nine or ten Also giving three odds to one or two Two to three or four And one to eight or nine So as to mark the smallest possible number of pictures And to give them the largest possible number of marks Answer Ten pictures Twenty-nine marks Arranged thus in a three by ten matrix The first row has crosses in the first nine columns And an odd in the tenth column The second row has crosses in the first five columns An empty entry in the sixth column And odds in columns seven to ten The third row has crosses in the first two columns And odds in the last eight columns Solution By giving all the crosses possible Putting into brackets the optional ones We get ten pictures marked thus in a three by ten matrix The first row has crosses in all columns The cross in the last column is put in brackets The second row has crosses in the first five columns The last cross is put in brackets And the columns six to ten are empty The third row has crosses in the first three columns The last cross is put in brackets And the columns four to ten are empty By then assigning odds in the same way Beginning at the other end We get nine pictures marked thus in a three by nine matrix The first row starts with seven empty columns Has an odd put in brackets in the eighth column And an odd in the ninth column The second row starts with five empty columns Has an odd put in brackets in the sixth column And odds in columns seven to nine The third row has nine odds The odd in the first column is put in brackets 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 ten necessary marks in the first row And in the third, but only seven in the second Hence we erase all optional marks In the first and third rows But let them stand in the second Twenty-two answers have been received Of these eleven give no working So in accordance with what I announced In my last review of answers I leave them unnamed Merely mentioning that five are right and six wrong Of the eleven answers With which some working is supplied Three are wrong C.H. begins with the rush assertion That under the given conditions The sum is impossible For he or she adds These initial correspondence Are dismally vague beings to deal with Perhaps it would be a better pronoun Ten is the least possible number of pictures Granted, therefore we must either give Two crosses to six Or two odds to five Why must, oh alphabetical phantom It is nowhere ordained that every picture Must have three marks Fifi sends a folio page of solution Which deserves the better fate She offers three answers In each of which ten pictures are marked With thirty marks In one she gives two crosses to six pictures In another to seven In the third she gives two odds to five Thus in every case ignoring the conditions I pause to remark that the condition Two crosses to four or five pictures Can only mean either to four or else to five If, as one competitor holds It might mean any number not less than four The words or five would be superfluous IEA, I am happy to say That none of these bloodless phantoms appear This time in the class list Is it idea with a D left out Gives two crosses to six pictures She then takes me to task For using the word ought Instead of not No doubt to one who thus rebels Against the rules laid down for her guidance The word must be distasteful But does not IEA remember the parallel case of adder That creature was originally a nether Then the two words took to banding The poor N backwards and forwards Like a shuttlecock The final state of the game being an adder May not a nod have similar become an odd Anyhow, odds and crosses is a very old game I don't think I ever heard it called Nods and crosses In the following class list I hope the solitary occupant of three Will sheath 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 meager that Like the nursery tale of Jack Minery I trust IEA will be merciful to the spelling It is scarcely to be distinguished from zero Class list First Guy Old cat Sea breeze Second Iyer Bradshaw of the future Effley H. Vernon Third Cat End of chapter 15 Chapter 16 of A Tangled Tale This LibriVox recording is in the public domain Recording by Abaii in October 2009 A Tangled Tale by Lewis Carroll Chapter 16 Answers to Nods 6 Problem 1 A and B began the year with only 1,000 pounds apiece They borrowed nought, they stole nought On the next New Year's Day they had 60,000 pounds 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 Edelpate makes them borrow zero and steal zero And uses both ciphers by putting them At the right-hand end of the 1,000 pounds Thus producing 100,000 pounds Which is well over the mark But, or to express it in Latin Atspes infracta has solved it even more ingenuously With the first cipher she turns the one of the 1,000 pounds into a nine And adds the results to the original sum Thus getting 10,000 pounds And in this, by means of the other zero She turns the one into a six Thus hitting the exact 60,000 pounds Class List First, Atspes infracta Second, Edelpate Problem 2 L makes five scarves While M makes two Z makes four While L makes three Five scarves of Z's weigh one of L's Five of M's weigh three of Z's One of M's is as warm as four of Z's And one of L's as warm as three of M's Which is best, giving equal weight in the results 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 five to two Z's to L's in the ratio of four to three In order to get one set of three 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 one, two-fifths, two-thirds 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 five to one Thus the marks for lightness are One-fifth, two-thirds, one And similarly, the marks for warmth Are three, one, one-fourth To get a total result Multiply L's three marks together And do the same for M and for Z The final numbers are One times one-fifth times three Two-fifths times two-thirds times one Two-thirds times one, times one-quarter That is, three-fifths, two-thirds, one-third That is, multiplying throughout by fifteen Which will not alter the proportion Nine, ten, five 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 ten feet long Ten wide, two deep Let B dig one six feet long Five wide, ten deep The cubical contents are two hundred, three hundred That is, B is best digger in the ratio of three to two Now try marking for length, width and depth separately Giving a maximum mark of ten to the best in each contest And then adding the results Of the twenty formula factors, 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 one He, she or it is adepate 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 wrongdoers Are Ire, Brecher of the Future, Fersbush And Pollux, who send a joint answer Grace Dead, Guy, Old Hen and Simple Susan The letter was once best of all The Old Hen has taken advantage of her simplicity And begulled 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 Grecia, M.M., Old Cat and R.E.X Tis Grace, 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 Ire, who also appears among the quite tutu Clifton C, F.B., Fifi, Grig, Janet And Mrs. Siri Gemp F.B. has not fallen into the common era She multiplies together the proportionate numbers she gets But in getting them she goes wrong By reckoning warmth as a demerit Possibly she is freshly burned or comes from Bombay Janet and Mrs. Siri Gemp have also avoided this era The method they have adopted is shrouded in mystery I scarcely feel competent to criticize it Mrs. Gemp says If Zuzu makes four while Lolo makes three Zuzu makes six while Lolo makes five Bad reasoning While Mimi makes two From these she concludes Therefore Zuzu excels in speed by one That is when compared with Lolo But what about Mimi? She then compares the three kinds of excellence Measured on this mystic scale Janet takes the statement that Lolo makes five while Mimi makes two To prove that Lolo makes three While Mimi makes one and Zuzu four Worse reasoning than Mrs. Gemp's And then concludes that Zuzu excels in speed by one-eighth Janet should have been able to line Mystery of mysteries The fourth set actually put Mimi at the top Arranging them as Mimi, Zuzu, Lolo They are Markian Coe Maltrep SBB First initial scarcely legible Maybe meant for J And Stanza The fifth set consists of an ancient fish and camel These ill-assorted comrades by dinner, food 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 fish-like ideas As to how numbers represent merit She says Lolo gains two and a half on Mimi Two and a half, what? Fish, fish, aren't thou in thy duty? Of the five winners I put Belbus and the Elder Traveler slightly below the other three Belbus for defective reasoning The other for scanty working Belbus 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 First Dynamite EBDL Joram Second Belbus DL the Traveler With regard to not five I beg to express to this inertia 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 odds and crosses Should have caused them to waste so much time on 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 This inertia would have been in the first class If she had sent in the solution she now offers End of Chapter 16 Chapter 17 of A Tangled Tale This lip-revox recording is in the public domain Recording by Avayee in October 2009 A Tangled Tale by Lewis Carroll Chapter 17 Answers to not seven Problem Given that one glass of lemonade, three sandwiches And seven biscuits cost one and two pence And that one glass of lemonade, four sandwiches And ten biscuits cost one and five pence Find the cost of one, a glass of lemonade, a sandwich And a biscuit and two, two glasses of lemonade Three sandwiches and five biscuits Answer, one, eight pence Two, one and seven pence Solution, this is best treated algebraically Let X equal the cost in pence of a glass of lemonade Y of a sandwich and Z of a biscuit Then we have X plus three Y plus seven Z Equals 14 And X plus four Y plus ten Z Equals 17 And we require the values of X plus Y plus Z And of two X plus three Y plus five 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 two of the three 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 third 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 first equation from the second Which eliminates lemonade and gives Y plus three Z equals three Or Y equals three minus three Z And then substituting this value of Y in the first Which gives X minus two Z equals five That is X equals five plus two Z Now, if we substitute these values of X, Y In the quantities whose values are required The first becomes the quantity five plus two Z Plus the quantity three minus three Z Plus Z, that is eight And the second becomes two times the quantity five plus two Z Plus three times the quantity three minus three Z Plus five Z, that is nineteen Notice the answers are one, eight pence, two, one and seven pence 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 improving 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 twenty-six competitors who have sent in What I may call accidental solutions Had had a question to deal with where every number contained eight or ten digits I suspect it would have been a case of Silvered is the raven hare, see, patience Before any solution would have been hit on by the most ingenious of them Forty-five answers have come in, of which forty-four 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 ten of the thirteen cases, the answer is right Of the remaining twenty-eight, no less than twenty-six Have sent in accidental solutions And therefore fall short of the highest honors 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, one sandwich plus three biscuits Equals three pence, then followed the amount of the unknown Builds, 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, this inertia has sent in a piece of incorrect working Peruse the horrid details and shudder She takes eggs, call it Y As the cost of a sandwich, and concludes rightly enough That a biscuit will cost the quantity three minus Y over three She then subtracts the second equation from the first And deduces three Y plus seven times the quantity three minus Y over three Minus four Y plus ten times the quantity three minus Y Divided by three equals three By making two mistakes in this line, she brings out Y equals two over two Try it again, or viz inertia Away with inertia, infuse a little more viz And you will bring out the correct, though uninteresting result Zero equals zero This will show you that it is hopeless to try to coax Any one of these three unknowns to reveal its separate value The other competitor who is wrong throughout Is either JMC or TMC But whether he be a juvenile miscalculator Or a true mathematician confused He makes the answers seven pence and one and five pence He assumes with too much confidence That biscuits were half a penny each And that Clara paid for eight, though she only ate seven We will now consider the thirteen who's 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 one sandwich and three biscuits cost three pence And proceeds therefore one sandwich equals one and a half pence Three biscuits equals one and a half pence One lemonade equals six pence Dynamite begins like Anita and then proves rightly That a biscuit costs less than a penny When she concludes wrongly that it must cost half a penny FCW 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 aren't the extenuating circumstances? She begins by assuming that lemonade is four pence a glass And sandwiches three pence each Making with the two given equations Four conditions to be fulfilled by three miserable unknowns And, having naturally developed this into a contradiction She then tries five pence and two pence with a similar result Not a bene This process might have been carried on through the whole of the tertiary period Without gratifying one single mega theorem 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 plus three biscuits she makes equal to four Four what? Mayfair makes the astounding assertion that the equation s plus three b equals three Is evidently only satisfied by s equals two halves b equals one half Old Cat believes that the assumption that a sandwich costs one and a half pence 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 OVL is beyond my comprehension He takes the given equations as one and two Then, by the process two minus one deduces rightly Equation three, that is s plus three b equals three And then, again, by the process times three, a hopeless mystery deduces three s plus four b equals four I have nothing to say about it, I give it up C Breeze says it is immaterial to the answer Why? In what proportion three pence is divided between the sandwich and the three biscuits So she assumes s equals one and a half pence b equals half a penny Stanza is one of a very irregular metro At first, she, like Janet, identifies sandwiches with biscuits She then tries two assumptions s equals one, b equals two thirds and s equals one half, b equals two sixths and naturally ends in contradictions Then she returns to the first assumption and finds the three unknowns separately Quote est absurdum Stiletto identifies sandwiches and biscuits as articles Is the word ever used by confectioners? I fancy it What is the next article, ma'am? Was limited to linen drapers Two sisters first assume that biscuits are for a penny And then that there are two 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 Payet What is a turtle payet, please? And old Crow who sent a joint answer And YY adopt the same method YY gets the equation s plus 3b equals 3 And then says this sum must be apportioned in one of the three following ways It may be, I grant you, but YY do you say must? I fear it is possible for YY to be too wise 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, O mysterious birds Of those who win honors, the Shetland snark must have the third class alter himself He has only answered half the question, that is, the amount of Clara's luncheon The two little old ladies he pitilessly lives in the midst of their difficulty I beg to assure him, with thanks for his friendly remarks And entrance fees and subscriptions are things unknown in that most economical of clubs, the not-untires 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 a second class in sections, according to the number of steps The two kings are fearfully deliberate Those walking quick or taking shortcuts is inconsistent with kingly dignity But really, in reading Thezius' 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 Cole into Cole King Coelos, or Coel, seems to have reigned soon after Arthur's time Henry of Huntington identifies him with the king Coel, who first built Walter on Colster, which was named after him In the Chronicle of Robert of Gloucester we read After King Irirag, of whom we have, he told, Marius's son was king, Coel to Monon bold And his son was after him, Coel was his name Both he it were Coel to men, end 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 pence and 19 pence 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 lady's 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 two buns, one queen cake, two sausage rolls and a bottle of zoedome Total, one and nine pence, and one bun, two queen cakes, a sausage roll and a bottle of zoedome Total, one and four pence Suppose Clara's unknown luncheon to have been three buns, one queen cake, one sausage roll and two bottles of zoedome While the two little sisters had been indulging in eight buns, four queen cakes, two sausage rolls and six bottles of zoedome Poor souls, how thirsty they must have been If Balbus will kindly try this by his principle of two assumptions Just assuming that a bun is one penny and a queen cake two pence and then that a bun is three pence and a queen cake three pence 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 two pence each, the queen cakes three pence, the sausage rolls six pence and the zoedome two pence a bottle So that Clara's third luncheon had cost one and seven pence and her thirsty friends had spent four and four pence Another remark of Balbus I will quote and discuss, for I think that it also may yield a moral for some of my readers 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 veil victim with it till it has reached a 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 dropped and you appealed 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 checkers and who should I meet but build sikes, 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 darbys, he says There's your arithmetical policeman Now try the other method I see somebody running but he was well gone or ever I got neither place So I just took a look around in the garden and I noticed the foot marks where the chap had come right across your flower beds There was good big foot marks, surely, and I noticed as the left foot went down the hill 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 rubbed my hand on the wall where he got over and there was suit 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 promiscuous and I says, it's build sikes, 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 Heckler and Maltrep are the only two who have used the methods certain either to produce the answer or else to prove it impossible So they must share between them the highest honors Class List First Heckler Maltrep Second Paragraph 1 Two Steps Adelaide Clifton C EKC Guy Lanconu Little Jack Nil Desperandum Simple Susan Yellow Hammer Woolly One Paragraph 2 Three Steps AA A Christmas Carol Afternoon Tea An Appreciative School Ma'am Baby Belbys Bog Oak The Red Queen Wallflower Paragraph 3 Four Steps Hawthorne Durham SSG Paragraph 4 Five Steps A Stepney Coach Paragraph 5 Six Steps Bay Laurel Bradshaw of the Future Paragraph 6 Nine Steps Old King Cole Paragraph 7 Fourteen Steps Phezeos Answers to Correspondence I have received several letters on the subjects of knots 2 and 6, which led me to think some further explanation desirable. In knot 2 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. Trojanus, however, 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 6, the first problem was, of course, a mere jeu de mot, 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, said he, 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 faulted 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 not six 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 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, A and B, who are to compete in French, German and Italian. A. Let it be laid down that the result is to depend on their relative knowledge of each subject, so that, whether their marks for French be one, two or one hundred, two hundred, the result will be the same. And let it also be laid down that, if they get equal marks on two papers, the final marks are to have the same ratio as those of the third paper. This is a case of ordinary double rule of three. We multiply A's three marks together and do the same for B. Note that, if A gets a single zero, his final mark is zero, even if he gets full marks for two papers, while B gets only one mark for each paper. This of course would be very unfair on A, though a correct solution under the given conditions. B. The result is to depend, as before, on relative knowledge, 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 multiply it as in A, and so much nearer that it would be necessary to use the other multipliers twice to produce the same result as in A. That is, if the French ratio were two tenths and the other's two ninths, one ninth, so that the ultimate ratio by method A would be two over forty-five, I should multiply instead by two thirds, one third, giving the result one third which is nearer to two tenths than if he had used method A. C. The result is to depend on actual amount of knowledge of the three subjects collectively. Here we have to ask two questions. One, what is to be the unit, that is, standard to measure by, in each subject? Two, 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 one hundred. All lower amounts are represented by numbers between zero and one hundred. Then, if these units are to be of equal value, we simply add A's three marks together and do the same for B. 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 zero 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 add A'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 not six was meant to belong to a variety A, and to make this clear I inserted the following passage. Usually the competitors differ in one point only. Saas, last year, Fifi and Gogo met the same number of scars in the Trialvik, 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 to Balbus, 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 to Fifi, who says, I think a little more consideration will show you that our era of adding the proportional numbers together for each candidate instead of multiplying is no era at all. Why, even if addition had been 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 are accustomed to indulging when criticizing the answer, that such a tone must not be, be not, agreeable to the person's concern to 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 anyone, and that I sincerely regret any annoyance I may have thus inadvertently given. May I hope that in future they will recognize the distinction between severe language used in sober earnest and the words of unmeant bitterness, which Colleridge 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 a 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 demure. I hold that to use language likely to annoy any of my correspondence would not be in the least justified by the plea that I was quite certain of being correct. I trust that the not untyres and I are not on such terms as those. I beg to thank G.B. for the offer of a puzzle, which, however, is too like the old one, make four nines into one hundred. Chapter 18 Answers to not eight Paragraph one The peaks Problem? Place 24 peaks in four styes so that, as you go round and round, you may always find the number in each stye nearer to 10 than the number in the last. Answer Place eight peaks in the first stye, ten in the second, nothing in the third, and six in the fourth. Ten is nearer ten than eight, nothing is nearer ten than ten, six is nearer ten than nothing, and eight is nearer ten than six. This problem is noticed by only two correspondents. Balbus says, it certainly cannot be solved mathematically, nor do I see how to solve it by any verbal quibble. Nolan's volance makes her radiancy change the direction of going round, and even then is obliged to add, the peaks must be carried in front of her. Paragraph two, the Gramstiffs. Problem? Paragraphs start from a certain point, both ways, every 15 minutes. A traveller, starting on foot along with one of them, meets one in 12 and a half minutes. When will he be overtaken by one? Answer In six minutes one quarter. 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 two and a half minutes, it goes in that time as far as the traveller walks in 12 and a half. That is, it goes five times as fast. Now the overtaking omnibus is A behind the traveller when he starts, and therefore goes A plus X, while he goes X. Hence A plus X equals five X, that is, four X equals A, and X equals A over four. This distance would be traversed by an omnibus in 15 over four minutes, and therefore by the traveller in five times 15 over four. Hence he is overtaken in 18 minutes three quarters after starting, that is, in six minutes one quarter after meeting the omnibus. Four answers have been received, of which two are wrong. Dynamite rightly states that the overtaking omnibus reached a point where they met the other omnibus five minutes after they left. But wrongly concludes that, going five times as fast, it would overtake them in another minute. The travellers are five minutes walk ahead of the omnibus and must walk one fourth of this distance farther before the omnibus overtakes them, which will be one-fifth of the distance traversed by the omnibus in the same time. This will require one minute, one quarter more. Nolens-Volens tries it by process like Achilles and the tortoise. He rightly states that when the overtaking omnibus leaves the gate, the travellers are one-fifth of A ahead. And that it will take the omnibus three minutes to traverse this distance. During which time the travellers, he tells us, go one-fifteenth of A, this should be one-twenty-fifth. The traveller is being now one-fifteenth of A ahead. He concludes that the work remaining to be done is for the travellers to go one-sixtieth of A while the omnibus goes one-twelfth. The principle is correct and might have been applied earlier. Class List First Belbus Delta End of Chapter 18 Chapter 19 of A Tangled Tail This LibriVox recording is in the public domain. Recording by Abahi in October 2009. A Tangled Tail by Lewis Carroll. Chapter 19 Answers to not nine. Paragraph one. 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 these places 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. Heckler says 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 to Heckler, 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 underwater. Magpie says 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. Alt King Cole takes the same view as Heckler. Timpanum and Windex assume 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 lands themselves or rather sets themselves floating in the same boat as Heckler. I regret that there is no class list to publish for this problem. Paragraph 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, etc., which series has no end. He concludes that the water will rise without limit. Is this true? Solution? No, this series can never reach four inches since, however many terms we take, we are always short of four inches by an amount equal to the last term taken. Three answers have been received, but only two seem to be worthy of honors. Timpanum says that the statement about the stick is merely a blind to which the old answer may well be applied, solviture ambulando or rather mergendo. I trust Timpanum will not test this in his own person by taking the place of the Man in Balbus' Essay. He would infallibly be drowned. Old King Cole rightly points out that the series 2, 1, etc., is a decrease in geometrical progression, while Windex rightly identifies the fallacy as that of Achilles and the tortoise. Class List? First, Old King Cole, Windex. Paragraph 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 dimension of the garden. Answer? 60, 60 and a half. 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. That is, if x be the width, x times the quantity x plus one half equals 3,630. Solving this quadratic, we find x equals 60. Hence, the dimensions are 60, 60 and a half. 12 answers have been received, 7 right and 5 wrong. CGL, Neybop, Old Crow and Timpanum assume 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. CGL's working consists of dividing 3,630 by 60. Whence came this divisor or seagull? Divination? Or was it a dream? I fear this solution is worth nothing. Old Crow's is shorter and so, if possible, worth rather less. He says the answer is at once seen to be 60 by 60 and a half. Neybop's calculation is short, but as rich as a Neybop in era. He says that the square root of 3,630 multiplied by two equals the length plus the breadth. That is, 60.25 times two equals 120 and a half. His first assertion is only true of a square garden. His second is irrelevant, since 60.25 is not the square root of 3,630. Neybop, this will not do. Timpanum says that by extracting the square root of 3,630 we get 60 yards with a remainder of 30 over 60 or half a yard, which we add so as to make the oblong 60 by 60 and a half. This is very terrible, but worse remains behind. Timpanum proceeds 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 center a small plot of ground, two yards long by half a yard wide, the only space not occupied by walk. But Belbus expressly said that the walk used up the whole of the area. Oh, Timpanum, my Timpa is exhausted, my brain is numb. I can say no more. Heckler indulges again and again in that most fatal of all habits in computation. The making two mistakes which cancel each other. She takes x as the width of the garden in yards and x plus one half as its length and makes her first coil the sum of x and a half, x and a half, x minus one, x minus one, that is 4x minus three. But the fourth term should be x minus one and a half so that her first coil is half a yard too long. Her second coil is the sum of x minus two and a half, x minus two and a half, x minus three, x minus three. Here the first term should be x minus two and the last x minus three and a half. These two mistakes 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 the end of 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, dynamite, jenet, magpie and taffy make the same assumption as CGL and Co. They then solve by a quadratic. Magpie also tries it by arithmetical progression but fails to notice that the first and last coils have special values. Alumnus Etonae attempts to prove what CGL assumes by a particular instance, taking a garden six by five and a half. He ought to have proved that generally what is true of one number is not always true of others. Old King Cole solves it by an arithmetical progression. It is right but too lengthy to be worth as much as a quadratic. Windex proves it very neatly by pointing out that the 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 yard in one direction and then turns a right angle and goes half yard in another direction. Class List First Windex Second Alumnus Etonae Old King Cole Third Dynamite Janet Magpie Taffy End of Chapter 19 Chapter 20 of A Tangled Tale This LibriVox recording is in the public domain. Recording by Avayee in October 2009. A Tangled Tale by Lewis Carroll. Chapter 20 Answers to not 10 Paragraph 1 The Chelsea Pensioners Problem If 70% have lost an eye, 75% an ear, 80% an arm, 85% a leg, what percentage at least must have lost all four? Answer 10 Solution I adopt that of Polar Star as being better than my own. Adding the wounds together we get 70 plus 75 plus 80 plus 85 equals 310 among 100 men which gives 3 to each and 4 to 10 men. Therefore the least percentage is 10. 19 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. Janet makes it 35 and 2 tenths. I am sorry she has misunderstood the question and has supposed that those who had lost an ear were 75% of those who had lost an eye and so on. Of course on this supposition the percentages must all be multiplied together. This she has done correctly but I can give her no honors as I do not think the question will fairly bear her interpretation. 3 score and 10 makes it 19 and 2 eighths. 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% I suppose. Dividing by 4 she gets 77 and a half as average percentage. Again dividing by 4 she gets 19 and 2 eighths as percentage wounded four times. Does she suppose wounds of different kinds to absorb each other so to speak? Then no doubt the data are 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 be transferable so that 2 fourths of these unfortunates can obtain perfect health by handing over their wounds to the remaining one fourth? Granting these suppositions her answer is right or rather if the question had been A road is covered with one inch of gravel along 77 and a half percent of it. How much of it could be covered four inches deep with the same material? Her answer would have been right but alas that wasn't the question. Delta makes the most amazing assumptions. Let everyone who has not lost an eye have lost an ear. Let everyone who has not lost both eyes and ears have lost an arm. Her ideas of a battlefield 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 considers possible. Next come eight writers who have made the unwarrantable assumption that because 70% have lost an eye therefore 30% have not lost one so that they have both eyes. This is illogical. If you give me a bag containing 100 sovereigns and if in an hour I come to you, my face not beaming with gratitude nearly so much as when I receive 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 signs of this illogical octagon are as follows in alphabetical order. Algonon Bray, Dynamite, GSC, Jane E, JDW, Magpie, who makes the delightful remark therefore 90% have two of something recalling to one's memory that fortunate Monarch with whom Circuses was so much pleased that he gave him ten of everything. SSG and Tokyo. Bradshaw of the future and T.R. do the question in a piecemeal fashion. On the principle that the 70% and the 75% though commenced at opposite ends of the 100 must overlap by at least 45% 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. Dynamite, GSC, Jane E, JDW, Magpie, SSG, Tokyo. Paragraph two, change of day. I must postpone, synodier 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? Paragraph three, the son's 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 be x, y, the quantity x plus y. Now, if a plus b equals 2c then the quantity a minus n plus the quantity b minus n equals two times the quantity c minus n, whatever be the value of n. Hence the second relationship if ever true was always true. Hence it was true at first. But it cannot be true that x and y are together double of the quantity x plus y. Hence it must be true of the quantity x plus y together with x or y and it does not matter which we take. We assume then the quantity x plus y plus x equals 2y. That is y equals 2x. Hence the three ages were at first x, 2x, 3x and the number of years since that time is two thirds of 6x. That is, is 4x. Hence the present ages are 5x, 6x, 7x. The ages are clearly integers since this is only the year when one of my sons comes of age. Hence 7x equals 21, x equals 3 and the other ages are 15, 18. 18 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 there were 14, 11 and 8. As a Roman father I ought to withhold the name of the rush writer, but respect for age makes me break the rule. It is 3 score and 10. Jane E. also asserts that the ages at first were 9, 6, 3. Then she calculates the present ages, leaving the second occasion unnoticed. Old hen is nearly as bad. She tried various numbers till I found one that fitted all the conditions, but merely scratching up the earth and pecking about is not the way to solve a problem or venerable bird. And close after old hen prowls with hungry eyes, old cat, who calmly assumes to begin with that the son who comes of age is the eldest. Eat your bird, Puss, for you will get nothing from me. There are yet two zeros to dispose of. Minerva assumes that on every occasion 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 future says, let the ages at first be 9, 6, 3, then assumes that the second occasion was six years afterwards and then these baseless assumptions brings out the right answers. Guide future travellers and thou wilt, thou art no Bradshaw for this age. Of those who win honors, they merely honourable are two. Dynamite ascertains rightly the relationship between the three ages at first, but then assumes one of them to be six, thus making the rest of her solution tentative. MFC does the algebra all write up to the conclusion that the present ages are 5Z, 6Z and 7Z, it then assumes without giving any reason that 7Z equals 21. Of the more honourable, delta attempts a novelty to discover which son comes of age by elimination. It assumes successively that it is the middle one and that it is the youngest and in each case it apparently brings out an absurdity. Still, as the proof contains the following bit of algebra, 63 equals 7x plus 4y, therefore 21 equals x plus 4 sevenths of y, I trust it will admit that its proof is not quite conclusive. The rest of its work is good. Magpie betrays the deplorable tendency of her tribe to appropriate any stray conclusion she comes across without having any strict logical right to it. Assuming A, B, C as the ages at first and D as the number of the years that have elapsed since then, she finds rightly the three equations 2A equals B, C equals B plus A, D equals 2B. She then says supposing that A equals 1, then B equals 2, C equals 3 and D equals 4. Therefore, for A, B, C, D, four numbers are wanted which shall be to each other as 1 to 2 to 3 to 4. It is not therefore that I detect the unconsciousness of this bird. The conclusion is true, but this is only because the equations are homogeneous, that is 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 plus 1 equals B, B plus 1 equals C, supposing A equals 1, then B equals 2 and C equals 3. Therefore, for A, B, C, three numbers are wanted which shall be to one another as 1 to 2 to 3. Would you not flutter down into it, O magpie, as amiable as a dove? Simple Susan is anything but simple to me. After ascertaining that the three ages at first are as 3 to 2 to 1, she says, then as two-thirds of their sum added to one of them equals 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 plus one age equals 21. Therefore, sum plus three-halves of one age equals 31 and a half. But three-halves of one age cannot be less than one and a half. Here I perceive that simple Susan would on no account present a guinea to a newborn 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 five possible sets of ages and eliminates four of them. Suppose that instead of five, there had been five million possible sets. Would simple Susan have courageously ordered in the necessary gallon of ink and ream of paper? The solution sent in by CR is, like that of simple Susan, partly tentative and so does not rise higher than being clumsily right. Among those who have earned the highest honors, Algonom Bray solves the problem quite correctly but adds that there is nothing to exclude the supposition that all the ages were fractional. This would make the number of answers infinite. Let me meekly protest that I never intended my readers to devote the rest of their lives to writing out answers. E. M. Riggs points 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 sugar is 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 that year, need not have done so by that day so that he might be only twenty. This gives a second solution, that is twenty, twenty-four, twenty-eight. Well said, pure crystal. Verily, thy fair discourse hath been as sugar. Class List First, Algonom Bray, an old phogy, E. M. Riggs, G. S. C., S. S. G., Tokyo, T. R., White Sugar. Second, C. R., Delta, Magpie, Simple Susan. Third, Dynamite, M. F. C. I have received more than one remonstrance on my assertion in the Chelsea-Pensioner's problem that it was illogical to assume from the datum seventy percent have lost an eye that thirty percent have not. Algonom Bray states, as a parallel case, suppose Tommy's father gives him four apples and he eats one of them. How many has he left? And says, I think we are justified in answering three. I think so too. There is no must here, and the data evidently meant to fix the answer exactly. But if the question was set me, how many must he have left, I should understand the data to be that his father gave him four at least, but may have 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 meter of an antique song is beyond my compass, and my puppets were neither distinctly in my life, like those I now address, nor yet, like Alice and a mock turtle, distinctly out of 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 moral.