795] 
NUMBERS. 
595 
3. The only numbers divisible by a number JS T = a a b^c y ... are the numbers 
a a '№c y •••> where each exponent a is equal to or greater than the corresponding 
exponent a. And conversely the only numbers which divide A are those of the form 
a a 'Wc y '..., where each index a' is at most equal to the corresponding index a; and 
in particular each or any of the indices a! may be = 0. Again, the least common 
multiple of two numbers A = a a №c y ... and A' = a a '№'c y ' ... is a a "№"c y "..., where each 
index a' is equal to the largest of the corresponding indices a, a ;—observe that any 
one or more of the indices a, /3, y,..., a', /3', y,..., may be = 0, so that the theorem 
extends to the case where either of the numbers A r , A', has prime factors which are 
not factors of the other number. And so the greatest common measure of two 
numbers A — a a №c y ... and A' = a a 'b^'c y '... is a a "Wc y '..., where each index a" is equal 
to the least of the corresponding indices a and a. 
4. The divisors of JST=a a №c y ... are the several terms of the product 
(1 + a + ... + a a ) (1 + b + ... + ¥) (1 + c + ... + c y ), 
where unity and the number N itself are reckoned each of them as a divisor. Hence 
the number of divisors is = (a + 1) (/3 +1) (7 + 1) ..., and the sum of the divisors is 
_ (q«+ 1 - 1) (№ +1 - 1) (c y+1 - 1) ... 
(a-l)(6-l)(c-l)... 
5. In ii= a a №c y ... the number of integers less than N and prime to it is 
To find the numbers in question write down the series of numbers 1, 2, 3,..., N; 
strike out all the numbers divisible by a, then those divisible by b, then those divisible 
by c, and so on; there will remain only the numbers prime to A r . For actually 
finding the numbers we may of course in striking out those divisible by b disregard 
the numbers already struck out as divisible by a, and in striking out with respect to 
c disregard the numbers already struck out as divisible by a or b, and so on; but 
in order to count the remaining numbers it is more convenient to ignore the previous 
strikings out. Suppose, for a moment, there are only two prime factors a and b, then 
the number of terms struck out as divisible by a is = N. —, and the number of 
CL 
terms struck out as divisible by b is = A. ^; but then each term divisible by ab will 
have been twice struck out; the number of these is =N.^r, and thus the number 
u ~ ab’ 
which is = N 
of the remaining terms is Ail 
a b ab. 
treating in like manner the case of three or more prime factors a, b, c,... we arrive at the 
general theorem. The formula gives <£ (1) = 1 viz. when N = 1, there is no factor 1 - i; 
and it is necessary to consider 0(1) as being =1. The explanation is that 0(A) 
75—2