BISHOP TERROT ON THE SUMS OF THE DIGITS OF NUMBERS. 89 
Prop. IV. 
If a have a common divisor with n—1 as v, then paand ga will have the same 
n—1 
ultimate sum if g—p= 3 

Let pa=P.n—1+,r, therefore ga=pa+"—* a=P.n—1 ere tole. But v isa 
divisor of a; therefore ga=P.n—1+7r+6n—1=P,n»—147, that is, pa and qa have 
same ultimate sum. 
Prop. V. 
If a be a divisor of »—1, or nat =o, then paand qa will have the same ul- 
timate sum if g—p=v. 
Let pa=P.n—1+r, ga=pat+va=Pn—1l+r+n—1=(P4+1)nm—1+4r. 
Prop. VI. 
If P=Q+R. The ultimate sum of P = ultimate sum of (sum Q + sum R). 
Let Q=mn—1+r, R=m,.n—1 +7, 
P=Q+R=mim,n—l+r+r, 
But 7 and 7, being single digits, their aggregate is either a single digit, or 
n—1+ asingle digit. In the former case, the ultimate sum of P = sum of Q + 
sum of R. In the latter, sum of P = sum (sum of Q + sum of R). 
Cor. If R be a multiple of n—1, or r,=n—1, sum of P = sum of Q. 
Prop. VII. 
From these propositions it follows, that in any arithmetical series, whose 
common difference is prime to n—1, the ultimate sum of any term (the pt) = the 
ultimate sum of (p+g¢n—1)™: but that no two terms at any other interval can 
have the same ultimate sum; and hence, that all the terms from the p® to the 
(p+n—1)" range, as to their ultimate sums, through all the digits of the scale. 
For if the p™ term =sn2—1+7, then the (p+g¢n—1)" term =s.n—1+rign—1.6 
= st+ogn—1 +P 
Again, let p™ term=a, g=a+g—p.; but by Prop. I., since 6 is now taken 
prime to n—1, and g—p is neither n—1, nor a multiple of it, the g™ term must 
have an ultimate sum different from the p*. 
Ex. 1. The successive multiples of any number prime to 9, are an arithmetic 
series whose common difference is that number. Thus, the multiples of 5 are, 
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, Ke. 
Sums 1G, 2, tnd, s, 4,9, 5, ¥, 6 de. 
