*Answer*.---Ten.

*Solution*.---(I adopt that of POLAR 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." JANET makes it "35 and ths." 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 has done correctly, but I can give her no honours, as I do not think the question will fairly bear her interpretation. THREE SCORE AND TEN makes it "19 and ths." 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 of all of it. She makes the number of "pensionerswounded once" to be 310 ("pre cent.," I suppose!): dividing by 4, she gets 77 and a half as "average percentage:" again dividing by 4, she gets 10 and ths. 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 ths of these unfortunates can obtain perfect health by handing over their wounds to the remaining th? 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 per cent. of it. How much of it ould be covered 4 inches deep with the same material?" her answer *would* have been right. But alas, that *wasn't* the question! DELTA makes 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 considers *possible*.

Next come eight writers who have made the unwarrantable assumption that, because 70 per cent. have lost an eye, *therefore* 30 per cent. have *not* lost one, so that they have *both* eyes. This is illogical. If you give me a bag of 100 sovereigns, and if in an hour I come back to you (my face not beaming 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., and TOKIO.

BRADSHAW OF THE FUTURE and 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 by *at least* 45 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 placed in the first class, without my composing a Triumphal Ode for each!

OLD CAT. | POLAR STAR. | |

OLD HEN. | SIMPLE SUSAN. | |

WHITE SUGAR. |

BRADSHAW OF THE FUTURE. | T. R. |

ALGERNON BRAY. | J. D. W. | |

DINAH MITE. | MAGPIE. | |

G. S. C. | S. S. G. | |

JANE E. | TOKIO. |

*Answer*.---"15 and 18"

*Solution*.-Let the ages at first be `x`, `y`, (`x`+`y`). Now if `a`+`b` = 2`c`, then (`a`-`n`) + (`b`-`n`) = 2(`c`-`n`), whatever be the value of of `n`. Hence the second relationship, if *ever* true, was *always* true. Hence it was true at first. But it cannot be that `x` and `y` are together double of (`x`+ `y`). Hence it must be true of (`x`+ `y`) together with `x` or `y`; and it does not matter which we take. We assume, then, (`x`+ `y`) + `x` = 2`y`, *i.e.* `y` = 2`x`. Hence the three ages were, at first, `x`, 2`x`, 3`x`; and the number of years, since that time is two-thirds of 6`x`, `i.e.` is 4`x`. Hence the present ages are 5`x`, 6`x`, 7`x`. The ages are clearly *integers*, since this is only "the year when one of my sons comes of age." Hence 7`x`=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, I *ought* to withhold the name of the rash writer; but respect for age makes me break the rule: it is THREE AND TEN. 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, oh 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 zeroes 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 the second occasion was 6 years afterwards, and on these baseless assumptions brings out the right answers. Guide *future* travellers, an thou wilt: thou art no Bradshaw for *this* Age!

Of those who win honours, the merely "honourable" are two. DINAH MITE ascertains (rightly) the relationship between the three ages at first, but then *assumes* one 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 5`z`, 6`z`, and 7`z`; it then assumes, without giving any reason, that 7`z` = 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 = 7`x` + 4`y`; 21 = `x` + 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 tendency of the 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 years that have elapsed since then, she finds (rightly) the 3 equations, 2`A` = `B`, `C` = `B` + `A`, `D` = 2`B`. She then says "supposing that `A` = 1, then `B` = 2, `C` = 3, and `D` = 4. Therefore for `A`, `B`, `C`, `D`, four numbers are wanted which shall be to each other as 1 : 2 : 3 : 4." It is in the "therefore" that I detect the unconscientiousness of this bird. The conclusion *is* true, 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`; supposing `A` = 1, then `B` = 2, and `C` = 3. *Therefore* for `A`, `B`, `C`, three numbers are wanted which shall be to one another as 1 : 2 : 3," would you not flutter down into it, oh MAGPIE, as amiably as a Dove? SIMPLE SUSAN is anything but simple to *me*. 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 one age, = 31 and a half. But 3 halves of one age cannot be less than 1 and-a-half (here I perceive that SIMPLE SUSAN would 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 ahd been 5 million possible sets? Would SIMPLE SUSAN have courageously ordered in the necessary gallon of ink and ream of paper?

The solution sent in by C.R. 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 honours, ALGERNON 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. RIX 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 20. This gives a second solution, viz., 20, 24, 28. Well said, pure Crystal! Verily, thy "fair discourse has been as sweet as sugar"!

ALGERNON BRAY. | S. S. G. | |

AN OLD FOGEY. | TOKIO. | |

E. M. RIX. | T. R. | |

G. S. C. | WHITE SUGAR. |

C. R. | MAGPIE. | |

DELTA. | SIMPLE SUSAN. |

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 the

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 distinctly *in* my life (like those I now address), nor yet (like Alice and the Mock Turtle) distinctly *out* of it. Yet let me at least fancy, as I lay down my 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.

Title | Epigraph | Preface |
Contents |

Knot | I II III IV V VI VII VIII IX X |

Answers | I II III IV V VI VII VIII IX X |

Stefan Bilaniuk, Department of Mathematics, Trent University

Maintained by Stefan Bilaniuk. Last updated 1998.08.22.