is an integer, whence ([Prop. 2]) b must divide 10ⁿ, and so must all the divisors of b. If, then, among the divisors of b there be any prime numbers except 2 and 5, we have a prime number (which is of course a number prime to 10) not dividing 10, but dividing one of its powers, which is absurd.
Prop. 4. If b be prime to a, all the multiples of b, as b, 2b, ... up to (a-1)b must leave different remainders when divided by a. For if, m being greater than n, and both less than a, we have mb and nb giving the same remainder, it follows that mb-nb, or (m-n)b, is divisible by a; whence ([Prop. 2]), a divides m-n, a number less than itself, which is absurd.
If a number be divided into its prime factors, or reduced to a product of prime numbers only (as in 360 = 2 × 2 × 2 × 3 × 3 × 5), and if a, b, c, &c. be the prime factors, and α, β, γ, &c. the number of times they severally enter, so that the number is aα × bᵝ × cᵞ × &c., then this can be done in only one way: For any prime number v, not included in the above list, is prime to a, and therefore to aα, to b and therefore to bᵝ and therefore to aα × bᵝ Proceeding in this way, we prove that v is prime to the complete product above, or to the given number itself.
The number of divisors which the preceding number aαbᵝcᵞ ... can have, 0 and itself included, is (α + 1)(β+ 1)(γ + 1).... For aα as the divisors 1, a, a² ... aα and no others, α + 1 in all. Similarly, bᵝ has β+ 1 divisors, and so on. Now as all the divisors are made by multiplying together one out of each set, their number (page 202) is (α + 1)(β + 1)(γ+ 1)....
If a number, n, be divisible by certain prime numbers, say 3, 5, 7, 11, then the third part of all the numbers up to n is divisible by 3, the fifth part by 5, and so on. But more than this: when the multiples of 3 are omitted, exactly the fifth part of those which remain are divisible by 5; for the fifth part of the whole are divisible by 5, and the fifth part of those which are removed are divisible by 5, therefore the fifth part of those which are left are divisible by 5. Again, because the seventh part of the whole are divisible by 7, and the seventh part of those which are divisible by 3, or by 5, or by 15, it follows that when all those which are multiples of 3 or 5, or both, are removed, the seventh part of those which remain are divisible by 7; and so on. Hence, the number of numbers not exceeding n, which are not divisible by 3, 5, 7, or 11, is ¹⁰/₁₁ of ⁶/₇ of ⁴/₅ of ²/₃ of n. Proceeding in this way, we find that the number of numbers which are prime to n, that is, which are not divisible by any one of its prime factors, a, b, c, ... is
| n | a -1 | b - 1 | c - 1 | ... | ||
| a | b | c |
or aα-1bβ-1cγ-1 ... (a - 1)(b - 1)(c - 1)....
Thus, 360 being 2³3²5, its number of divisors is 4 × 3 × 2, or 24, and there are 2³3.1.2.4 or 96 numbers less than 360 which are prime to it.
Prop. 5. If a be prime to b, then the terms of the series, a, a², a³, ... severally divided by b, must all leave different remainders, until 1 occurs as a remainder, after which the cycle of remainders will be again repeated.