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_, (_x_ + _y_). Now, if _a_
+ _b_ = 2_c_, then (_a_-_n_) + (_b_-_n_) = 2(_c_-_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 (_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 SCORE 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 that 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_; [** therefore] 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 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 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;
[** therefore] 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 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 had 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 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 the
_datum_ "70 p. c. have lost an eye," that 30 p. c. have _not_. ALGERNON
BRAY states, 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 the _data_ are evidently meant to fix the answer
_exactly_: but, if the question were set me "how many _must_ he have
left?", I should understand the _data_ to be that his father gave him 4
_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 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 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
LONDON: RICHARD CLAY AND SONS, PRINTERS.
[TURN OVER.
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* * * * *
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* * * * *
Transcriber's note
The following changes have been made to the text:
Page 88: "he corners of the" changed to "the corners of the".
Page 95: "Aix-le-Bains" changed to "Aix-les-Bains".
Page 108: "3/5, 2, 1/3" changed to "3/5, 2/3, 1/3".
Page 114: "10 of the 12 cases" changed to "10 of the 13 cases".
Page 121: "four-and fourpence" changed to "four-and-fourpence".
Last page: "Fifth Thousand" changed to "Fifty Thousand".