BISHOP TERROT ON THE SUMS OF THE DIGITS OF NUMBERS. 91 
To illustrate this and some of the succeeding propositions, I shall here 
introduce a table of successive powers of digits prime to 9, with their ultimate 
sums. 
Powers 2, 4, 8, 16, 32, 64, 128, &c. 
Base 2. Sums recur after 6 terms. 
Sums” 2 4 Se erts Ly 2) s&e 
Powers 4, 16, 64, 256, 1024, &e. 
Base 4 Sums recur after 3 terms. 
PS ho Ay oe wikis) SCs 

Base 5 | POWeS 5 25. 125, 625, 8125, 15625, 78125, &e. | Sums recur after 
Sms oO 7, 8, 40 °.2; 1, 5, &e. 6 terms. 

Powers 7, 49, 343, 2401, 16807, &c. 
Base 7 
sums 7, 4, 1, 7% 4, &e. 
Sums recur after 3 terms. 
Powers 8, 64, 512, 4096, &c. 
Base 8 
Spams 8° 1, 8, “ih Ge. 
| sums recur after 2 terms. 

‘In this table, we may observe that in every case the sum of the digits recurs, 
but at different intervals. Next, if we take two complementary bases, as 5 and 4, 
_we find in the lines expressing the sums, that the first terms are respectively 
4 and 5, the 2d terms 7 and 7, the 3d 1 and 8, and so on; as was proved gene- 
rally in the last proposition. Lastly, we may observe that the digits 3, 6, 9, that 
is n—1, and the digits having a common divisor with n—1, never occur among the 
sums. It remains, then, for us to point out the reason of this last mentioned 
fact, and to discover the principle which fixes the period of recurrence. 
Prop. X. 
Every power of a number prime to n—1, must have the sum of its digits also 
prime to x—1. 
Let m, which is prime to n—1, be reduced to its prime factors, or let 
m=a.b.c, &¢. | 
Then m” = (a’.t.c? &e.) Here m” has no possible divisors except a, 6, ¢, &c., 
and by hyp. none of these are divisors of n—1, therefore m” is prime to »—1. 
Now let m” =gn—1+,r. Here gn=1 contains all the divisors of n—1. If, 
therefore, 7 contains any of those divisors g#—1 +7, or m” contains such divi- 
VOL. XVI. PART II. 2A 
