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the bean climbing spirally as at A above, whereas the French bean, or scarlet runner, the variety clearly selected by the artist in the absence of any authoritative information on the point, always climbs as shown at B. Very few seem to be aware of this curious little fact. Though the bean always insists on a sinistrorsal growth, as B, the hop prefers to climb in a dextrorsal manner, as A. Why, is one of the mysteries that Nature has not yet unfolded.

426.—THE HYMN-BOARD POSER.

This puzzle is not nearly so easy as it looks at first sight. It was required to find the smallest possible number of plates that would be neces- sary to form a set for three hymn-boards, each of which would show the five hymns sung at any particular service, and then to discover the lowest possible cost for the same. The hymn-book contains 700 hymns, and therefore no higher number than 700 could possibly be needed.

Now, as we are required to use every legiti- mate and practical method of economy, it should at once occur to us that the plates must be painted on both sides; indeed, this is such a common practice in cases of this kind that it would readily occur to most solvers. We should also remember that some of the figures may possibly be reversed to form other figures; but as we were given a sketch of the actual shapes of these figures when painted on the plates, it would be seen that though the 6's may be turned upside down to make 9's, none of the other figures can be so treated.

It will be found that in the case of the figures 1, 2, 3, 4, and 5, thirty-three of each will be required in order to provide for every possible emergency; in the case of 7, 8, and 0, we can only need thirty of each; while in the case of the figure 6 (which may be reversed for the figure 9) it is necessary to provide exactly forty- two.

It is therefore clear that the total number of figures necessary is 297 ; but as the figures are painted on both sides of the plates, only 149 such plates are required. At first it would appear as if one of the plates need only have a number on one side, the other side being left blank. But here we come to a rather subtle point in the problem.

Readers may have remarked that in real life it is sometimes cheaper when making a purchase to buy more articles than we require, on the principle of a reduction on taking a quantity : we get more articles and we pay less. Thus, if we want to buy ten apples, and the price asked is a penny each if bought singly, or ninepence a dozen, we should both save a penny and get two apples more than we wanted by buying the full twelve. In the same way, since there is a regular scale of reduction for plates painted alike, we actually save by having two figures painted on that odd plate. Supposing, for example, that we have thirty plates painted alike with 5 on one side and 6 on the other. The rate would be 4¾d., and the cost 11s. 10½d. But if the odd plate with, say, only a 5 on one side of it have a 6 painted on the other side, we get thirty-one plates at the reduced rate of 4½d., thus saving a farthing on each of the previous thirty, and reducing the cost of the last one from is. to 4½d.

But even after these points are all seen there comes in a new difficulty : for although it will be found that all the 8's may be on the backs of the 7's, we cannot have all the 2's on the backs of the 1's, nor all the 4's on the backs of the 3's, etc. There is a great danger, in our attempts to get as many as possible painted alike, of our so adjusting the figures that some particular combination of hynms cannot be represented.

Here is the solution of the difficulty that was sent to the vicar of Chumpley St. Winifred. Where the sign × is placed between two figmres, it implies that one of these figures is on one side of the plate and the other on the other side.

							d.		£	s.	d.
31	plates painted	5	×	9	@	4	½	=	0	11	7	½
30	plates painted„	7	×	8	@	4	¾	=	0	11	10	½
21	plates painted„	1	×	2	@	7		=	0	12	3
21	plates painted„	3	×	0	@	7		=	0	12	3
12	plates painted„	1	×	3	@	9	¼	=	0	9	3
12	plates painted„	2	×	4	@	9	¼	=	0	9	3
12	plates painted„	9	×	4	@	9	¼	=	0	9	3
8	plates painted„	4	×	0	@	10	¼	=	0	6	0
1	plates painted„	5	×	4	@	12		=	0	1	0
1	plates painted„	5	×	0	@	12		=	0	1	0
–––
149 plates @ 6d. each								=	3	14	6
									£7	19	1

Of course, if we could increase the number of plates, we might get the painting done for nothing, but such a contingency is prevented by the condition that the fewest possible plates must be provided.