Complete Works of Lewis Carroll (96 page)

ANSWERS TO KNOT IV.

Problem.
—"There are 5 sacks, of which Nos.
1, 2, weigh 12 lbs.; Nos.
2, 3, 13½ lbs.; Nos.
3, 4, 11½ lbs.; Nos.
4, 5, 8 lbs.; Nos.
1, 3, 5, 16 lbs.
Required the weight of each sack."

Answer.
—"5½, 6½, 7, 4½, 3½."

The sum of all the weighings, 61 lbs., includes sack No.
3
thrice
and each other
twice
.
Deducting twice the sum of the 1st and 4th weighings, we get 21 lbs.
for
thrice
No.
3,
i.e.
, 7 lbs.
for No.
3.
Hence, the 2nd and 3rd weighings give 6½ lbs., 4½ lbs.
for Nos.
2, 4; and hence again, the 1st and 4th weighings give 5½ lbs., 3½ lbs., for Nos.
1, 5.

Ninety-seven answers have been received.
Of these, 15 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 the 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 this "mute inglorious" band are Common Sense, D.
E.
R., Douglas, E.
L., Ellen, I.
M.
T., J.
M.
C., Joseph, Knot I, 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 neatness 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 knell of all my hopes!"
Why, oh why, did you assume that the 4th and 5th bags weighed 4 lbs.
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 Ardmore, A ready Reckoner, Arthur, Bog-Lark, Bog-Oak, Bridget, First Attempt, J.
L.
C., M.
E.
T., Rose, Rowena, Sea-Breeze, Sylvia, Thistledown, Three-Fifths Asleep, Vendredi, and Winifred.
Bog-Lark tries it by a sort of "rule of false," assuming experimentally that Nos.
1, 2, weigh 6 lbs.
each, and having thus produced 17½, instead of 16, as the weight of 1, 3, and 5, 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 seemed perfectly clear" to her that, "3 out of the 5 sacks being weighed twice over,
2
⁄
5
of 45 = 27, must be the total weight of the 5 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 No.
1 weighed 5½ lbs.
The rest all do it, wholly or partly, by guess-work.

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 I.
those whose answers seemed specially short and neat, and into Class III.
those that seemed specially long or clumsy.
Of this last set, A.
C.
M., Furze-Bush, James, Partridge, R.
W., and Waiting for the Train, have sent long wandering solutions, the substitutions having no definite method, but seeming to have been made to see what would come of it.
Chilpome and Dublin Boy omit some of the working.
Arvon Marlborough Boy only finds the weight of
one
sack.

CLASS LIST

I.

B.
E.
D.

C.
H.

Constance Johnson.

Greystead.

Guy.

Hoopoe.

J.
F.
A.

M.
A.
H.

Number Five.

Pedro.

R.
E.
X.

Seven Old Men.

Vis Inertiæ.

Willy B.

Yahoo.

II.

American Subscriber.

An appreciative schoolma'am.

Ayr.

Bradshaw of the Future.

Cheam.

C.
M.
G.

Dinah Mite.

Duckwing.

E.
C.
M.

E.
N.
Lowry.

Era.

Euroclydon.

F.
H.
W.

Fifee.

G.
E.
B.

Harlequin.

Hawthorn.

Hough Green.

J.
A.
B.

Jack Tar.

J.
B.
B.

Kgovjni.

Land Lubber.

L.
D.

Magpie.

Mary.

Mhruxi.

Minnie.

Money-Spinner.

Nairam.

Old Cat.

Polichinelle.

Simple Susan.

S.
S.
G.

Thisbe.

Verena.

Wamba.

Wolfe.

Wykehamicus.

Y.
M.
A.
H.

III.

A.
C.
M.

Arvon Marlborough Boy.

Chilpome.

Dublin Boy.

Furze-Bush.

James.

Partridge.

R.
W.

Waiting for the Train.

ANSWERS TO KNOT V.

Problem.
—To mark pictures, giving 3 x's to 2 or 3, 2 to 4 or 5, and 1 to 9 or 10; also giving 3 o's to 1 or 2, 2 to 3 or 4 and 1 to 8 or 9; so as to mark the smallest possible number of pictures, and to give them the largest possible number of marks.

Answer.
—10 pictures; 29 marks; arranged thus:—

 x

 x

 x

 x

 x

 x

 x

 x

 x

 o

 x

 x

 x

 x

 x

 o

 o

 o

 o

x

x

o

o

o

o

o

o

o

o

Solution.
—By giving all the x's possible, putting into brackets the optional ones, we get 10 pictures marked thus:—

 x

 x

 x

 x

 x

 x

 x

 x

 x

(x)

 x

 x

 x

 x

(x)

 x

 x

(x)

By then assigning o's in the same way, beginning at the other end, we get 9 pictures marked thus:—

(o)

 o

(o)

 o

 o

 o

(o)

 o

 o

 o

 o

 o

 o

 o

 o

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 10 necessary marks in the 1st row, and in the 3rd; but only 7 in the 2nd.
Hence we erase all optional marks in the 1st and 3rd rows, but let them stand in the 2nd.

Twenty-two answers have been received.
Of these 11 give no working; so, in accordance with what I announced in my last review of answers, I leave them unnamed, merely mentioning that 5 are right and 6 wrong.

Of the eleven answers with which some working is supplied, 3 are wrong.
C.
H.
begins with the rash assertion that under the given conditions "the sum is impossible.
For," he or she adds (these initialed correspondents are dismally vague beings to deal with: perhaps "it" would be a better pronoun), "10 is the least possible number of pictures" (granted): "therefore we must either give 2 x's to 6, or 2 o's to 5."
Why "must," oh alphabetical phantom?
It is nowhere ordained that every picture "must" have 3 marks!
Fifee sends a folio page of solution, which deserved a better fate: she offers 3 answers, in each of which 10 pictures are marked, with 30 marks; in one she gives 2 x's to 6 pictures; in another to 7; in the 3rd she gives 2 o's to 5; thus in every case ignoring the conditions.
(I pause to remark that the condition "2 x's to 4 or 5 pictures" can only mean "
either
to 4
or else
to 5": if, as one competitor holds, it might mean
any
number not less than 4, the words "
or
5" would be superfluous.) I.
E.
A.
(I am happy to say that none of these bloodless phantoms appear this time in the class-list.
Is it IDEA with the "D" left out?) gives 2 x's to 6 pictures.
She then takes me to task for using the word "ought" instead of "nought."
No doubt, to one who thus rebels against the rules laid down for her guidance, the word must be distasteful.
But does not I.
E.
A.
remember the parallel case of "adder"?
That creature was originally "a nadder": then the two words took to bandying the poor "n" backwards and forwards like a shuttlecock, the final state of the game being "an adder."
May not "a nought" have similarly become "an ought"?
Anyhow, "oughts and crosses" is a very old game.
I don't think I ever heard it called "noughts and crosses."

In the following Class-list, I hope the solitary occupant of III.
will sheathe 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 meagre that, like the nursery tale of "Jack-a-Minory" (I trust I.
E.
A.
will be merciful to the spelling), it is scarcely to be distinguished from "zero."

CLASS LIST.

I.

Guy.

Old Cat.

Sea-Breeze.

II.

Ayr.

Bradshaw of the Future.

F.
Lee.

H.
Vernon.

III.

Cat.