348 Answersin the order shown in Figure II, where you go round in a clockwise direction,
starting with I and jumping over a disc to the place for 2,jumping over another
for 3, and so on. Now to complete the star for the constant summation of 24,
as required, use this simple rule. To find H subtract the sum of Band C from
half the constant plus E. That is, subtract 6 from 15. We thus get 9 as the re-
quired number for H. Now you are able to write in successively IO at F (to
make 24), 6 at J, 12 at G, and 8 at K. There is your solution.You can write any five numbers you like in the pentagon, in any order, and
with any constant summation that you wish, and you will always get, by the
rule shown, the only possible solution for that pentagon and constant. But
that solution may require the use of repeated numbers and even negative
numbers. Suppose, for example, I make the pentagon I, 3, II, 7, 4, and the
constant 26, as in Figure III, then I shall find the 3 is repeated, and the repeated
4 is negative and must be deducted instead of added. You will also find that
if we had written our pentagon numbers in Figure II in any other order
we should always get repeated numbers.