92 BISHOP TERROT ON THE SUMS OF THE DIGITS OF NUMBERS. 
sors; but m” has been proved prime to x—1; therefore r contains no divisor of 
n—1, or is prime to it. 
Prop. XI. 
To determine the recurrence of ultimate sums in the series m, m?, m3, &c., 
m being any single digit. } 
If we can determine what term will have 1 for its ultimate sum, the problem 
is solved. For if m’ = pm—1+1, m’** = p’'n—1+™, or has same sum as first term, 
m'** —»’n—1+m?, and so on, or sums recur after g terms. 
Every number (m) is of the form 3p or 3p +1. 
1. If m be of form 3p, every power of m after the first is a multiple of 9, 
and consequently the sum of every power =9. 
2. If m be of form 3p+1, 
mi = 39M 4.9.3 pan clout ae +l * Sp +93pt1. 
In this expansion, every term is divisible by 9, except the two last, or 
m'=9s+3pq+l. 
Consequently m? will have 1 for its ultimate sum, if 3q = a multiple of 9; 
but since (m being one of the digits) p cannot =3, or a multiple of 3, g must. If, 
then, g=3, 3p q=9p, and the sum 1 will recur at every third term. 
3. If m=3p—1, m'=3p" — gp" . , . . =(8pq—1), the sign being — 
if g be even, + if ¢ be odd. . 
a. Let g be even. m’=9s—3pq+1; and this, as before, will give the ulti- 
mate sum 1, if 3 pq be a multiple of 9, or pg =a multiple of 3. If p be prime 
to 3, then g must be an even multiple of 3, as 6, 12, &c., or the ultimate sum 1 
recurs at every 6 terms. But if p be 3, or a multiple of 3, the sum will recur at 
every second term, for in that case g may be any even number. 
B. If g be odd, m’=9s+3pq—1, but by hyp. m?=97r+1, .. r—s.9=3pq—2, or 
3pq—2 is a multiple of 3, which is absurd. Therefore the sum 1 can never 
recur at an odd power, when 2 is of the form 3 p—1. 
If, now, we refer to the table given in Prop. IX., we see that of the bases 
- there employed, 4 and 7 are of the form 3 p+1, and in them the sum 1 recurs at 
every third term. 2, 5, and 8 are of the form 3p—1. In 2 and 5, p is prime to 
three, and therefore the sum 1 occurs at 6th term. In 8, p=3, and therefore the 
sum 1 recurs at every even term. 
Cor. Hence, if m be not a multiple of 3, 7. e. if it be prime to 9, m* has 1 for 
its ultimate sum, for 1 must occur at 2d, 3d, or 6th term, and 6 is a multiple of 
2 and 3; therefore in any case the 6th power must have 1 for its ultimate sum. 
