350 Answersmust equal the sum of the triangle DEF. This sum may be anything from 12
to 27 inclusive, except 14 and 25, which are impossible. We need only obtain
solutions for 12, 13, 15, 16, 17, 18, and 19, because from these all the comple-
mentaries, 27, 26, 24, 23, 22, 21, and 20, may be derived by substituting for
every number in the star its difference from 13.
(2) Every arrangement is composed of three independent diamonds,
AGHF, DKBL, and EMCI, each of which must always sum to 26.
(3) The sum of the numbers in opposite external triangles will always be
equal. Thus AIK equals LMF.
( 4) If the difference between 26 and the triangle sum ABC be added to any
number at a point, say A, it will give the sum of the two numbers in the rela-
tive positions of Land M. Thus (in Figure II) 10 + 13 = II + 12, and
6+13=8+11.
(5) There are six pairs summing to 13; they are 12 + I, II + 2, 10 + 3,
9 + 4, 8 + 5, 7 + 6, and one pair or two pairs may occur among the numbers
at the points, but never three. The relative positions of these pairs determine
the type of solution. In the regular type, as in Figure II, A and F and also G
and H, as indicated by the dotted lines, always sum to 13, though I subdivide
this class. Figures III and IV are examples of the two irregular types. There
are 37 solutions in all (or 74, if we count the complementaries described in my
first paragraph), of which 32 are regular and 5 irregular.
Of the 37 solutions, 6 have their points summing to 26. They are as follows:
10 6 2 3 I 4 7 9 5 12 II 8
9 7 4 3 2 6 II 5 10 12 85 4 6 8 2 I (^9 12 3) II 7 10
5 2 7 8 3 II 10 4 12 6 9
10 3 I 4 2 6 9 8 7 12 II 5
8 5 3 I 2 7 10 4 II 9 12 6
The first is our Figure II, and the last but one our Figure III, so a reference
to those diagrams will show how to write the numbers in the star. The reader
should write them all out in star form and remember that the 6 are increased
to 12 if you also write out their complementaries. The first four are of the
regular type and the last two of the irregular. If the reader should be tempted
to find all the 37 (or 74) solutions to the puzzle it will help him to know that,
where the six points sum to 24, 26, 30, 32, 34, 36, 38, the respective number
of solutions is 3, 6, 2, 4,7,6,9, making 37 in all.