Complete Works of Lewis Carroll (95 page)

Let A be No.
9, B No.
25, C No.
52, and D No.
73.

Then AB = √(12
2
+ 5
2
) = √169 = 13;

AC = 21;

AD = √(9
2
+ 8
2
) = √145 = 12 +

(N.B.
i.e.
"between 12 and 13.")

BC = √(16
2
+ 12
2
) = √400 = 20;

BD = √(3
2
+ 21
2
) = √450 = 21+;

CD = √(9
2
+ 13
2
) = √250 = 15+;

Hence sum of distances from A is between 46 and 47; from B, between 54 and 55; from C, between 56 and 57; from D, between 48 and 51.
(Why not "between 48 and 49"?
Make this out for yourselves.) Hence the sum is least for A.

Twenty-five solutions have been received.
Of these, 15 must be marked "0," 5 are partly right, and 5 right.
Of the 15, I may dismiss Alphabetical Phantom, Bog-Oak, Dinah Mite, Fifee, Galanthus Nivalis Major (I fear the cold spring has blighted our Snowdrop), Guy, H.M.S.
Pinafore, Janet, and Valentine with the simple remark that they insist on the unfortunate lodgers
keeping to the pavement
.
(I used the words "crossed to Number Seventy-three" for the special purpose of showing that
short cuts
were possible.) Sea-Breeze does the same, and adds that "the result would be the same" even if they crossed the Square, but gives no proof of this.
M.
M.
draws a diagram, and says that No.
9 is the house, "as the diagram shows."
I cannot see
how
it does so.
Old Cat assumes that the house
must
be No.
9 or No.
73.
She does not explain how she estimates the distances.
Bee's Arithmetic is faulty: she makes √169 + √442 + √130 = 741.
(I suppose you mean √741, which would be a little nearer the truth.
But roots cannot be added in this manner.
Do you think √9 + √16 is 25, or even √25?) But Ayr's state is more perilous still: she draws illogical conclusions with a frightful calmness.
After pointing out (rightly) that AC is less than BD she says, "therefore the nearest house to the other three must be A or C."
And again, after pointing out (rightly) that B and D are both within the half-square containing A, she says "therefore" AB + AD must be less than BC + CD.
(There is no logical force in either "therefore."
For the first, try Nos.
1, 21, 60, 70: this will make your premiss true, and your conclusion false.
Similarly, for the second, try Nos.
1, 30, 51, 71.)

Of the five partly-right solutions, Rags and Tatters and Mad Hatter (who send one answer between them) make No.
25 6 units from the corner instead of 5.
Cheam, E.
R.
D.
L., and Meggy Potts leave openings at the corners of the Square, which are not in the
data
: moreover Cheam gives values for the distances without any hint that they are only
approximations
.
Crophi and Mophi make the bold and unfounded assumption that there were really 21 houses on each side, instead of 20 as stated by Balbus.
"We may assume," they add, "that the doors of Nos.
21, 42, 63, 84, are invisible from the centre of the Square"!
What is there, I wonder, that Crophi and Mophi would
not
assume?

Of the five who are wholly right, I think Bradshaw Of the Future, Caius, Clifton C., and Martreb deserve special praise for their full
analytical
solutions.
Matthew Matticks picks out No.
9, and proves it to be the right house in two ways, very neatly and ingeniously, but
why
he picks it out does not appear.
It is an excellent
synthetical
proof, but lacks the analysis which the other four supply.

CLASS LIST.

Bradshaw of the Future

Caius.

Clifton C.

Martreb.

II.

Matthew Matticks.

III.

Cheam.

Crophi and Mophi.

E.
R.
D.
L.

Meggy Potts.

{Rags and Tatters.

{Mad Hatter.

A remonstrance has reached me from Scrutator on the subject of Knot I., which he declares was "no problem at all."
"Two questions," he says, "are put.
To solve one there is no data: the other answers itself."
As to the first point, Scrutator is mistaken; there
are
(not "is") data sufficient to answer the question.
As to the other, it is interesting to know that the question "answers itself," and I am sure it does the question great credit: still I fear I cannot enter it on the list of winners, as this competition is only open to human beings.

ANSWERS TO KNOT III.

Problem.
—(1) "Two travellers, starting at the same time, went opposite ways round a circular railway.
Trains start each way every 15 minutes, the easterly ones going round in 3 hours, the westerly in 2.
How many trains did each meet on the way, not counting trains met at the terminus itself?"
(2) "They went round, as before, each traveller counting as 'one' the train containing the other traveller.
How many did each meet?"

Answers.
—(1) 19.
(2) The easterly traveller met 12; the other 8.

The trains one way took 180 minutes, the other way 120.
Let us take the L.
C.
M., 360, and divide the railway into 360 units.
Then one set of trains went at the rate of 2 units a minute and at intervals of 30 units; the other at the rate of 3 units a minute and at intervals of 45 units.
An easterly train starting has 45 units between it and the first train it will meet: it does 2-5ths of this while the other does 3-5ths, and thus meets it at the end of 18 units, and so all the way round.
A westerly train starting has 30 units between it and the first train it will meet: it does 3-5ths of this while the other does 2-5ths, and thus meets it at the end of 18 units, and so all the way round.
Hence if the railway be divided, by 19 posts, into 20 parts, each containing 18 units, trains meet at every post, and, in (1), each traveller passes 19 posts in going round, and so meets 19 trains.
But, in (2), the easterly traveller only begins to count after traversing 2-5ths of the journey,
i.e.
, on reaching the 8th post, and so counts 12 posts: similarly the other counts 8.
They meet at the end of 2-5ths of 3 hours, or 3-5ths of 2 hours,
i.e.
, 72 minutes.

Forty-five answers have been received.
Of these 12 are beyond the reach of discussion, as they give no working.
I can but enumerate their names.
Ardmore, E.
A., F.
A.
D., L.
D., Matthew Matticks, M.
E.
T., Poo-Poo, and The Red Queen are all wrong.
Beta and Rowena have got (1) right and (2) wrong.
Cheeky Bob and Nairam give the right answers, but it may perhaps make the one less cheeky, and induce the other to take a less inverted view of things, to be informed that, if this had been a competition for a prize, they would have got no marks.
[N.B.—I have not ventured to put E.
A.'s name in full, as she only gave it provisionally, in case her answer should prove right.]

Of the 33 answers for which the working is given, 10 are wrong; 11 half-wrong and half-right; 3 right, except that they cherish the delusion that it was
Clara
who travelled in the easterly train—a point which the data do not enable us to settle; and 9 wholly right.

The 10 wrong answers are from Bo-Peep, Financier, I.
W.
T., Kate B., M.
A.
H., Q.
Y.
Z., Sea-Gull, Thistledown, Tom-Quad, and an unsigned one.
Bo-Peep rightly says that the easterly traveller met all trains which started during the 3 hours of her trip, as well as all which started during the previous 2 hours,
i.e.
, all which started at the commencements of 20 periods of 15 minutes each; and she is right in striking out the one she met at the moment of starting; but wrong in striking out the
last
train, for she did not meet this at the terminus, but 15 minutes before she got there.
She makes the same mistake in (2).
Financier thinks that any train, met for the second time, is not to be counted.
I.
W.
T.
finds, by a process which is not stated, that the travellers met at the end of 71 minutes and 26½ seconds.
Kate B.
thinks the trains which are met on starting and on arriving are
never
to be counted, even when met elsewhere.
Q.
Y.
Z.
tries a rather complex algebraical solution, and succeeds in finding the time of meeting correctly: all else is wrong.
Sea-Gull seems to think that, in (1), the easterly train
stood still
for 3 hours; and says that, in (2), the travellers met at the end of 71 minutes 40 seconds.
Thistledown nobly confesses to having tried no calculation, but merely having drawn a picture of the railway and counted the trains; in (1), she counts wrong; in (2) she makes them meet in 75 minutes.
Tom-Quad omits (1): in (2) he makes Clara count the train she met on her arrival.
The unsigned one is also unintelligible; it states that the travellers go "1-24th more than the total distance to be traversed"!
The "Clara" theory, already referred to, is adopted by 5 of these, viz., Bo-Peep, Financier, Kate B., Tom-Quad, and the nameless writer.

The 11 half-right answers are from Bog-Oak, Bridget, Castor, Cheshire Cat, G.
E.
B., Guy, Mary, M.
A.
H., Old Maid, R.
W., and Vendredi.
All these adopt the "Clara" theory.
Castor omits (1).
Vendredi gets (1) right, but in (2) makes the same mistake as Bo-Peep.
I notice in your solution a marvellous proportion-sum:—"300 miles: 2 hours :: one mile: 24 seconds."
May I venture to advise your acquiring, as soon as possible, an utter disbelief in the possibility of a ratio existing between
miles
and
hours
?
Do not be disheartened by your two friends' sarcastic remarks on your "roundabout ways."
Their short method, of adding 12 and 8, has the slight disadvantage of bringing the answer wrong: even a "roundabout" method is better than
that
!
M.
A.
H., in (2), makes the travellers count "one"
after
they met, not
when
they met.
Cheshire Cat and Old Maid get "20" as answer for (1), by forgetting to strike out the train met on arrival.
The others all get "18" in various ways.
Bog-Oak, Guy, and R.
W.
divide the trains which the westerly traveller has to meet into 2 sets, viz., those already on the line, which they (rightly) make "11," and those which started during her 2 hours' journey (exclusive of train met on arrival), which they (wrongly) make "7"; and they make a similar mistake with the easterly train.
Bridget (rightly) says that the westerly traveller met a train every 6 minutes for 2 hours, but (wrongly) makes the number "20"; it should be "21."
G.
E.
B.
adopts Bo-Peep's method, but (wrongly) strikes out (for the easterly traveller) the train which started at the
commencement
of the previous 2 hours.
Mary thinks a train, met on arrival, must not be counted, even when met on a
previous
occasion.

The 3, who are wholly right but for the unfortunate "Clara" theory, are F.
Lee, G.
S.
C., and X.
A.
B.

And now "descend, ye classic Ten!"
who have solved the whole problem.
Your names are Aix-les-Bains, Algernon Bray (thanks for a friendly remark, which comes with a heart-warmth that not even the Atlantic could chill), Arvon, Bradshaw of the Future, Fifee, H.
L.
R., J.
L.
O., Omega, S.
S.
G., and Waiting for the Train.
Several of these have put Clara, provisionally, into the easterly train: but they seem to have understood that the data do not decide that point.

CLASS LIST.

Aix-les-Bains.

Algernon Bray.

Bradshaw of the Future.

Fifee.

H.
L.
R.

Omega.

S.
S.
G.

Waiting for the train.

II.

Arvon.

J.
L.
O.