(related to Problem: 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 necessary 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 legitimate 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 is 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 $4\frac 34$d., and the cost $11$s. $10\frac 12$d. But if the odd plate with, say, only a $5$ on one side of it has a $6$ painted on the other side, we get thirty-one plates at the reduced rate of $4\frac 12$d., thus saving a farthing on each of the previous thirty, and reducing the cost of the last one from $1$s. to $4\frac 12$d.
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 $4$ 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 hymns 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 figures, 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| $5 \times 9$ at| $4\frac 12 =$| $0$| $11$| $7\frac 12$ 30| "| $7 \times 8$ at| $4\frac 34 =$| $0$| $11$| $10\frac 12$ 21| "| $1 \times 2$ at| $7 =$| $0$| $12$| $3$ 21| "| $3 \times 0$ at| $7 =$| $0$| $12$| $3$ 12| "| $1 \times 3$ at| $9\frac 14 =$| $0$| $9$| $3$ 12| "| $2 \times 4$ at| $9\frac 14 =$| $0$| $9$| $3$ 12| "| $9 \times 4$ at| $9\frac 14 =$| $0$| $9$| $3$ 8| "| $4 \times 0$ at| $10\frac 14 =$| $0$| $6$| $10$ 1| "| $5 \times 4$ at| $12 =$| $0$| $1$| $0$ 1| "| $5 \times 0$ at| $12 =$| $0$| $1$| $0$ 149| plates at 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.
This puzzle appeared in Tit-Bits, and the following remarks, made by me in the issue for 11th December 1897, may be of interest.
The "Hymn-Board Poser" seems to have created extraordinary interest. The immense number of attempts at its solution sent to me from all parts of the United Kingdom and from several Continental countries show a very kind disposition amongst our readers to help the worthy vicar of Chumpley St. Winifred over his parochial difficulty. Every conceivable estimate, from a few shillings up to as high a sum as £$1,347,$ $10$s., seems to have come to hand. But the astonishing part of it is that, after going carefully through the tremendous pile of correspondence, I find that only one competitor has succeeded in maintaining the reputation of the Tit-Bits solvers for their capacity to solve anything, and his solution is substantially the same as the one given above, the cost being identical. Some of his figures are differently combined, but his grouping of the plates, as shown in the first column, is exactly the same. Though a large majority of competitors clearly hit upon all the essential points of the puzzle, they completely collapsed in the actual arrangement of the figures. According to their methods, some possible selection of hymns, such as $111,$ $112,$ $121,$ $122,211,$ cannot be set up. A few correspondents suggested that it might be possible so to paint the $7$'s that upside down they would appear as $2$'s or $4$'s; but this would, of course, be barred out by the fact that a representation of the actual figures to be used was given.
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