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," a.s.suming 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," a.s.sumes (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 workmans.h.i.+p 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 Cla.s.s I. those whose answers seemed specially short and neat, and into Cla.s.s 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 subst.i.tutions 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.

CLa.s.s 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.

w.i.l.l.y 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 a.s.signing 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.