makes the distance "23 miles," while her revolutionary companion puts it
at "27." GERTY VERNON says "they had to go 4 miles along the plain, and
got to the foot of the hill at 4 o'clock." They _might_ have done so, I
grant; but you have no ground for saying they _did_ so. "It was 7-1/2
miles to the top of the hill, and they reached that at 1/4 before 7
o'clock." Here you go wrong in your arithmetic, and I must, however
reluctantly, bid you farewell. 7-1/2 miles, at 3 miles an hour, would
_not_ require 2-3/4 hours. A NIHILIST says "Let _x_ denote the whole
number of miles; _y_ the number of hours to hill-top; [** therefore] 3_y_ =
number of miles to hill-top, and _x_-3_y_ = number of miles on the other
side." You bewilder me. The other side of _what_? "Of the hill," you
say. But then, how did they get home again? However, to accommodate your
views we will build a new hostelry at the foot of the hill on the
opposite side, and also assume (what I grant you is _possible_, though
it is not _necessarily_ true) that there was no level road at all. Even
then you go wrong.
You say
"_y_ = 6 - (_x_ - 3_y_)/6, ..... (i);
_x_/4-1/2 = 6 ..... (ii)."
I grant you (i), but I deny (ii): it rests on the assumption that to go
_part_ of the time at 3 miles an hour, and the rest at 6 miles an hour,
comes to the same result as going the _whole_ time at 4-1/2 miles an
hour. But this would only be true if the "_part_" were an exact _half_,
i.e., if they went up hill for 3 hours, and down hill for the other 3:
which they certainly did _not_ do.
The sixteen, who are partially right, are AGNES BAILEY, F. K., FIFEE, G.
E. B., H. P., KIT, M. E. T., MYSIE, A MOTHER'S SON, NAIRAM, A
REDRUTHIAN, A SOCIALIST, SPEAR MAIDEN, T. B. C., VIS INERTIAE, and YAK. Of
these, F. K., FIFEE, T. B. C., and VIS INERTIAE do not attempt the second
part at all. F. K. and H. P. give no working. The rest make particular
assumptions, such as that there was no level road--that there were 6
miles of level road--and so on, all leading to _particular_ times being
fixed for reaching the hill-top. The most curious assumption is that of
AGNES BAILEY, who says "Let _x_ = number of hours occupied in ascent;
then _x_/2 = hours occupied in descent; and 4_x_/3 = hours occupied on
the level." I suppose you were thinking of the relative _rates_, up
hill and on the level; which we might express by saying that, if they
went _x_ miles up hill in a certain time, they would go 4_x_/3 miles on
the level _in the same time_. You have, in fact, assumed that they took
_the same time_ on the level that they took in ascending the hill. FIFEE
assumes that, when the aged knight said they had gone "four miles in the
hour" on the level, he meant that four miles was the _distance_ gone,
not merely the rate. This would have been--if FIFEE will excuse the
slang expression--a "sell," ill-suited to the dignity of the hero.
And now "descend, ye classic Nine!" who have solved the whole problem,
and let me sing your praises. Your names are BLITHE, E. W., L. B., A
MARLBOROUGH BOY, O. V. L., PUTNEY WALKER, ROSE, SEA BREEZE, SIMPLE
SUSAN, and MONEY SPINNER. (These last two I count as one, as they send a
joint answer.) ROSE and SIMPLE SUSAN and CO. do not actually state that
the hill-top was reached some time between 6 and 7, but, as they have
clearly grasped the fact that a mile, ascended and descended, took the
same time as two level miles, I mark them as "right." A MARLBOROUGH BOY
and PUTNEY WALKER deserve honourable mention for their algebraical
solutions being the only two who have perceived that the question leads
to _an indeterminate equation_. E. W. brings a charge of untruthfulness
against the aged knight--a serious charge, for he was the very pink of
chivalry! She says "According to the data given, the time at the summit
affords no clue to the total distance. It does not enable us to state
precisely to an inch how much level and how much hill there was on the
road." "Fair damsel," the aged knight replies, "--if, as I surmise, thy
initials denote Early Womanhood--bethink thee that the word 'enable' is
thine, not mine. I did but ask the time of reaching the hill-top as my
_condition_ for further parley. If _now_ thou wilt not grant that I am a
truth-loving man, then will I affirm that those same initials denote
Envenomed Wickedness!"
CLASS LIST.
I.
A MARLBOROUGH BOY.
PUTNEY WALKER.
II.
BLITHE.
E. W.
L. B.
O. V. L.
ROSE.
SEA BREEZE.
{SIMPLE SUSAN.
{MONEY-SPINNER.
BLITHE has made so ingenious an addition to the problem, and SIMPLE
SUSAN and CO. have solved it in such tuneful verse, that I record both
their answers in full. I have altered a word or two in BLITHE'S--which I
trust she will excuse; it did not seem quite clear as it stood.
* * * * *
"Yet stay," said the youth, as a gleam of inspiration lighted up the
relaxing muscles of his quiescent features. "Stay. Methinks it matters
little _when_ we reached that summit, the crown of our toil. For in the
space of time wherein we clambered up one mile and bounded down the same
on our return, we could have trudged the _twain_ on the level. We have
plodded, then, four-and-twenty miles in these six mortal hours; for
never a moment did we stop for catching of fleeting breath or for gazing
on the scene around!"
"Very good," said the old man. "Twelve miles out and twelve miles in.
And we reached the top some time between six and seven of the clock. Now
mark me! For every five minutes that had fled since six of the clock
when we stood on yonder peak, so many miles had we toiled upwards on the
dreary mountainside!"
The youth moaned and rushed into the hostel.
BLITHE.
The elder and the younger knight,
They sallied forth at three;
How far they went on level ground
It matters not to me;
What time they reached the foot of hill,
When they began to mount,
Are problems which I hold to be
Of very small account.
The moment that each waved his hat
Upon the topmost peak--
To trivial query such as this
No answer will I seek.
Yet can I tell the distance well
They must have travelled o'er:
On hill and plain, 'twixt three and nine,
The miles were twenty-four.
Four miles an hour their steady pace
Along the level track,
Three when they climbed--but six when they
Came swiftly striding back
Adown the hill; and little skill
It needs, methinks, to show,
Up hill and down together told,
Four miles an hour they go.
For whether long or short the time
Upon the hill they spent,
Two thirds were passed in going up,
One third in the descent.
Two thirds at three, one third at six,
If rightly reckoned o'er,
Will make one whole at four--the tale
Is tangled now no more.
SIMPLE SUSAN.
MONEY SPINNER.
ANSWERS TO KNOT II.
Sec. 1. THE DINNER PARTY.
_Problem._--"The Governor of Kgovjni wants to give a very small dinner
party, and invites his father's brother-in-law, his brother's
father-in-law, his father-in-law's brother, and his brother-in-law's
father. Find the number of guests."
_Answer._--"One."
* * * * *
In this genealogy, males are denoted by capitals, and females by small
letters.
The Governor is E and his guest is C.
A = a
|
+------+-+----+
| | |
b = B D = d C = c
| | |
| +---++--+ +-+-+
| | | | | |
e = E | g = G |
F ========= f
Ten answers have been received. Of these, one is wrong, GALANTHUS
NIVALIS MAJOR, who insists on inviting _two_ guests, one being the
Governor's _wife's brother's father_. If she had taken his _sister's
husband's father_ instead, she would have found it possible to reduce
the guests to _one_.
Of the nine who send right answers, SEA-BREEZE is the very faintest
breath that ever bore the name! She simply states that the Governor's
uncle might fulfill all the conditions "by intermarriages"! "Wind of the
western sea," you have had a very narrow escape! Be thankful to appear
in the Class-list at all! BOG-OAK and BRADSHAW OF THE FUTURE use
genealogies which require 16 people instead of 14, by inviting the
Governor's _father's sister's husband_ instead of his _father's wife's
brother_. I cannot think this so good a solution as one that requires
only 14. CAIUS and VALENTINE deserve special mention as the only two who
have supplied genealogies.
CLASS LIST.
I.
BEE.
CAIUS.
M. M.
MATTHEW MATTICKS.
OLD CAT.
VALENTINE.
II.
BOG-OAK.
BRADSHAW OF THE FUTURE.
III.
SEA-BREEZE.
Sec. 2. THE LODGINGS.
_Problem._--"A Square has 20 doors on each side, which contains 21 equal
parts. They are numbered all round, beginning at one corner. From which
of the four, Nos. 9, 25, 52, 73, is the sum of the distances, to the
other three, least?"
_Answer._--"From No. 9."
* * * * *
[Illustration]
Let A be No. 9, B No. 25, C No. 52, and D No. 73.
Then AB = [** sqrt](12^{2} + 5^{2}) = [** sqrt]169 = 13;
AC = 21;
AD = [** sqrt](9^{2} + 8^{2}) = [** sqrt]145 = 12+
(N.B. _i.e._ "between 12 and 13.")
BC = [** sqrt](16^{2} + 12^{2}) = [** sqrt]400 = 20;
BD = [** sqrt](3^{2} + 21^{2}) = [** sqrt]450 = 21+;
CD = [** sqrt](9^{2} + 13^{2}) = [** sqrt]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 [** sqrt]169 + [** sqrt]442 + [** sqrt]130 = 741. (I suppose you
mean [** sqrt]741, which would be a little nearer the truth. But roots
cannot be added in this manner. Do you think [** sqrt]9 + [** sqrt]16 is
25, or even [** sqrt]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.
I.
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 26-1/2
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.
I.
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.
III.
F. LEE.
G. S. C.
X. A. B.
ANSWERS TO KNOT IV.
_Problem._--"There are 5 sacks, of which Nos. 1, 2, weigh 12 lbs.; Nos.
2, 3, 13-1/2 lbs.; Nos. 3, 4, 11-1/2 lbs.; Nos. 4, 5, 8 lbs.; Nos. 1, 3,
5, 16 lbs. Required the weight of each sack."
_Answer._--"5-1/2, 6-1/2, 7, 4-1/2, 3-1/2."
* * * * *
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 6-1/2 lbs., 4-1/2 lbs. for Nos. 2,
4; and hence again, the 1st and 4th weighings give 5-1/2 lbs., 3-1/2
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 17-1/2, 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, 3/5 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 5-1/2 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 INERTIAE.
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