56
GREEK NUMERICAL NOTATION
by a number of multiples of ten, each less than 100, or by
multiples of 100, each less than 1000), and so on. In the final
proposition (25), with which the introductory lemmas close,
the factors are of all three kinds, some containing units only,
others being multiples of 10 (less than 100) and a third set
being multiples of 100 (less than 1000 in each case). As
Pappus frequently says, the proof is easy ‘ in numbers ’;
Apollonius himself seems to have proved the propositions by
means of lines or a diagram in some form. The method is the
equivalent of taking the indices of all the separate powers of
ten included in the factors (in which process ten =10 1 counts
as 1, and 100 —10 2 as 2), adding the indices together, and then
dividing the sum by 4 to obtain the power of the myriad
(10000) which the product contains. If the whole number in
the quotient is n, the product contains (10000)” or the
n-myriad in Apollonius’s notation. There will in most cases
be a remainder left after division by 4, namely 3, 2, or 1 : the
remainder then represents (in our notation) 3, 2, or 1 more
ciphers, that is, the product is 1000, 100, or 10 times the
^-myriad, or the 10000”, as the case may be.
We cannot do better than illustrate by the main problem
which Apollonius sets himself, namely that of multiplying
together all the numbers represented by the separate letters
in the hexameter:
ApripiSos KXeTre Kpdros e£o\ov tvvea Kovpou.
The number of letters, and therefore of factors, is 38, of which
10 are multiples of 100 less than 1000, namely p, t, a, r, p, r,
o-, v, p (=100, 300, 200, 300, 100, 300, 200, 600, 400, 100),
17 are multiples of 10 less than 100, namely p, l, o, k, A, l, k, o, £,
o, o, v, v, v, k, o, l ( = 40, 10, 70, 20, 30, 10, 20, 70, 60, 70, 70, 50,
50, 50, 20, 70, 10), and 11 are numbers of units not exceeding
9, namely a, e, 8, e, e, a, e, e, e, a, a ( = 1, 5, 4, 5, 5, 1, 5, 5, 5, 1, 1).
The sum of the indices of powers of tqn contained in the
factors is therefore 10.2 + 17.1 =37. This, when divided by
4, gives 9 with 1 as remainder. Hence the product of all the
tens and hundreds, excluding the bases in each, is 10 . 10000 9 .
We have now, as the second part of the operation, to mul
tiply the numbers containing units only by the bases of all the
other factors, i.e. ^beginning with the bases, first of the hun
dreds, then of the tens) to multiply together the numbers: