There is a very large number of different ways in which arithmetical
signs may be placed between the nine digits, arranged in numerical
order, so as to give an expression equal to 100. In fact, unless the
reader investigated the matter very closely, he might not suspect that
so many ways are possible. It was for this reason that I added the
condition that not only must the fewest possible signs be used, but also
the fewest possible strokes. In this way we limit the problem to a
single solution, and arrive at the simplest and therefore (in this case)
the best result.
Just as in the case of magic squares there are methods by which we may
write down with the greatest ease a large number of solutions, but not
all the solutions, so there are several ways in which we may quickly
arrive at dozens of arrangements of the "Digital Century," without
finding all the possible arrangements. There is, in fact, very little
principle in the thing, and there is no certain way of demonstrating
that we have got the best possible solution. All I can say is that the
arrangement I shall give as the best is the best I have up to the
present succeeded in discovering. I will give the reader a few
interesting specimens, the first being the solution usually published,
and the last the best solution that I know.
Signs. Strokes.
1 + 2 + 3 + 4 + 5 + 6 + 7 + (8 × 9) = 100 ( 9 18)
- (1 × 2) - 3 - 4 - 5 + (6 × 7) + (8 × 9)
= 100 (12 20)
1 + (2 × 3) + (4 × 5) - 6 + 7 + (8 × 9)
= 100 (11 21)
(1 + 2 - 3 - 4)(5 - 6 - 7 - 8 - 9) = 100 ( 9 12)
1 + (2 × 3) + 4 + 5 + 67 + 8 + 9 =100 (8 16)
(1 × 2) + 34 + 56 + 7 - 8 + 9 = 100 (7 13)
12 + 3 - 4 + 5 + 67 + 8 + 9 = 100 (6 11)
123 - 4 - 5 - 6 - 7 + 8 - 9 = 100 (6 7)
123 + 4 - 5 + 67 - 8 - 9 = 100 (4 6)
123 + 45 - 67 + 8 - 9 = 100 (4 6)
123 - 45 - 67 + 89 = 100 (3 4)
It will be noticed that in the above I have counted the bracket as one
sign and two strokes. The last solution is singularly simple, and I do
not think it will ever be beaten.