93.—­THE MYSTIC ELEVEN.

Most people know that if the sum of the digits in the odd places of any number is the same as the sum of the digits in the even places, then the number is divisible by 11 without remainder.  Thus in 896743012 the odd digits, 20468, add up 20, and the even digits, 1379, also add up 20.  Therefore the number may be divided by 11.  But few seem to know that if the difference between the sum of the odd and the even digits is 11, or a multiple of 11, the rule equally applies.  This law enables us to find, with a very little trial, that the smallest number containing nine of the ten digits (calling nought a digit) that is divisible by 11 is 102,347,586, and the highest number possible, 987,652,413.

94.—­THE DIGITAL CENTURY.

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 x 9) = 100 ( 9 18)

— (1 x 2) — 3 — 4 — 5 + (6 x 7) + (8 x 9)
= 100 (12 20)

1 + (2 x 3) + (4 x 5) — 6 + 7 + (8 x 9)
= 100 (11 21)

(1 + 2 — 3 — 4)(5 — 6 — 7 — 8 — 9) = 100 ( 9 12)

    1 + (2 x 3) + 4 + 5 + 67 + 8 + 9 =100 (8 16)

    (1 x 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)