92 BISHOP TERROT ON THE SUMS OF THE DIGITS OF NUMBERS. 



sors ; but m*" has been proved prime to n-1 ; therefore r contains no divisor of 

 n— 1, or is prime to it. 



Pkop. XI. 



To determine the recuiTence of ultimate sums in the series m, m-, m', &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' = p.n—1 + 1, »«'*'= p'.n — l + in, or has same sum as first term, 

 m^*^=p".n—l + m-, and so on, or sums recur after q terms. 



Every number (m) is of the form 3/> or 3^ ± 1. 



1. If m be of form 3^^, 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 3 jo + 1, 



In this expansion, every term is divisible by 9, except the two last, or 



w'^Os + S/jy + l. 



Consequently m' will have 1 for its ultimate sum, if Sjo*/ = a multiple of 9; 

 but since (m being one of the digits) p cannot = 3, or a multiple of 3, q must. If, 

 then, (7=3, 3pq=9p, and the sum 1 will recm- at every third term. 



3. Itm=3p-1, m''=3p\'' - q.3p\''~^ .... =f (3 i» y-l), the sign being - 

 if 9 be even, + if §■ be odd. 



a. Let q be even. m''=9 s—3pq + l ; and this, as before, will give the ulti- 

 mate sum 1, if 3 pq he a multiple of 9, or pq = a multiple of 3. If p be prime 

 to 3, then q must be an even multiple of 3, as G, 12, &c., or the ultimate sum 1 

 recurs at every 6 terms. But if p be 3, or a multiple of 3, the sum wUl recur at 

 every second term, for in that case q may be any even number. 



p. It qhe oM, m''=9s + 3pq-1, but by hyp. w'' = 9r + l, .-. r^.9 = 3joy-2, or 

 3pq-2 is a multiple of 3, which is absurd. Therefore the sum 1 can never 

 recur at an odd power, when ?« is of the form 3/>-l. 



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 op + 1, and in them the sum 1 recurs at 

 every third term. 2, 5, and 8 are of the form 3p-l. 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. 



Con. Hence, if m be not a multiple of 3, i. 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. 



