We have now four series, the total number of points equal 15, as there ought to be, for one cube is absent.
Let us now take another example (see fig. 866), and by working as before we have four series again, viz:—
| 1st Series.—1, 7, 1 | (2) even. |
| 2nd Series.—2, 11, 3, 8, 4, 15, 2 | (6) even. |
| 3rd Series.—5, 12, 13, 5 | (3) uneven. |
| 4th Series.—9, 14, 10, 9 | (3) uneven. |
This gives us only 14 as a total, because 6 has not been touched at all.
And now for the rule, so that we may be able to ascertain in advance, when we have established our series, whether we shall find our puzzle right or wrong at the end. We must put aside all unplaced numbers and take no notice of uneven series. Only the even series must be regarded.
Thus if we do not find 1, or if we find 2, 4, or 6, the problem will come into A as a result. If we find 1, 3, 5, or 7, the case will eventuate as in B (fig. 864). Let us apply the rule to the problems we have worked, and then the reason will be apparent.
In the first we find three even series; the problem will then end as in B diagram (fig. 864), for the number of like series is odd.
In the second we find two even series (pairs); we shall find our problem work out as in diagram A (fig. 864), for the number of like series is even, one pair in each.
We are now in possession of a simple rule, both rapid and infallible, and which will save considerable trouble, as we can always tell beforehand how our puzzle will come out. Any one can test the practicability of the rules for himself, but we may warn the reader that he will never be able to verify every possible instance, for the possible cases are represented by the following sum—