90 BISHOP TERROT ON THE SUMS OF THE DIGITS OF NUMBERS. 
Ex. 2. But let the number whose multiples are taken have a common divisor 
with 9 (n—1) as 6. 
The series is 6, 12, 18, 24, 30, 36, 42, 48, 54, 60. 
Sums ~. .. (6,°3, 9; 6) a, Us me: 
Where the sums recur at every third term, because 6 and 9 have a common 
divisor 3, and = that is “— —* <3. (Prop. V.) 
Ex. 3. The recurrence of the sums, according to Prop. V., may be more 
strikingly illustrated, if we use a notation whose root is 13, and, consequently, 
n—1=12. If we express the successive multiples of 6 in this notation, we must 
adopt three additional characters for 10,11, 12. Let these be 1,,1,,1,. The 
successive multiples in this notation are, 
6,1,, 15, 11,, 24, 21,, 33, 39, 42, 48, &e. 
Sums . . 6,1,, 6, 1, 6, 1,, &c, where we see the sums recur after 

12 
two terms, because »—1=12, and Goze 
Prop. VIII. 
If x be even, and »—1 consecutive terms of an arithmetic series be taken, the 
ultimate sum of the digits of their aggregate is x—1. But if n be odd, the ulti- 
: Loeset —I.n—2 : . 
mate sum will be x—1, or sum ( on) according as b, the common differ- 
ence, is even or odd. 
n— = 

For aggregate of n—1 terms of arithmetic series = (2a+n—2.6) . 
2Gh =e tek ee. > If n be even sees is integer, whether } be even or odd. 
Therefore the latter term is a multiple of »—1, and, consequently, the whole 
a being a se of x—1, has n—1 for its ultimate sum (Prop. II., Cor.) 
ED: 

is integer only when 2 is even. 


PRoPOEX. 
If we assume as bases two numbers whose sum is s.x—1, and take a series 
of the successive powers of each, then of the two series expressing the sums of 
digits of successive powers, the even terms are identical, while the odd terms are 
complemental, that is, their sum is x—1. 
Let m+m,=s.n—1, m,=s.n—1—m 
m' = (sm—1) — p(s.n—1pP_ : Wes emt as p is even or 
Ait Here every term et the ee isa multiple of s.a—1 
Therefore if p be even, sum of m,= sum of m’. 
But if p be odd, sum of mn, = sum of (Q..—1)—sum of m =n—1— sum of m. 
