598 
PROBLEMS AND SOLUTIONS. 
[48 5 
Solution by Professor Cayley. 
1. In adding any two numbers, we carry a certain number of times; and it is 
easy to see that the sum of the digits of the two components, less the sum of the 
digits of the sum, is equal to nine times the number of carryings; moreover, that the 
number of carryings is equal to the number of borrowings, if either of the components 
be subtracted from the sum. 
2. The same thing is true in any scale of notation, only, instead of nine, we 
have the radix of the scale, less unity: say the theorem is 
S (to) 4- S (n) — S (to 4- n) = (p — 1) x. 
3. If p be a prime number, the number of times that the factor p occurs in 
II (to) is 
E 
w 
\pj 
+ E - + E — + &C., 
,p 
P 
„ to 
Avhere E ( —) denotes the integer part of —, and similarly E —-) &c. the integer part 
of —, &c. ; the series is, of course, finite. 
P 
p 
\p 
Hence the number of times that the factor p occurs in * s 
tf = E(™—) + e( ——) + &c. -E(-)-e(—)i - &c. -El-)- e(~) - &c. 
\ P J \ p 2 J \pJ \p-J \p/ VpV 
4. Hence, expressing m, n, m + n in the scale to the radix p, suppose 
m — a + bp + cp 2 + dp 3 , n = a' + b'p + c'p- + d'p 3 , m + n = a + ¡3p + yp 2 + Bp 3 , 
we have 
E 
+ &c. = b + cp + dp 2 + c + dp + d = d (p- + p + 1) + c (p + 1) + b ; 
and similarly for 
whence 
E - +&c., E 
\p 
TO + n 
+ &c. 
(p - 1) N = B (p 3 - 1) + 7 (p 2 - 1) + ß (p - 1) 
— d (p 3 — 1) — c (p 2 — 1) — b (p — 1) 
- d! (p 3 - 1) - c'(p 2 - 1) — b'{p — 1) 
= {to + n — S (to + ?i)| — {to — S (to)} — [n 
= S (to) + S (n) — S (to + n), = (p — 1) x, 
S(n)}