What are the divisors of 9?

1, 3, 9

There is a total of 3 positive divisors.
The sum of these divisors is 13.
The arithmetic mean is 4.3333333333333.

3 odd divisors

1, 3, 9

How to compute the divisors of 9?

A number N is said to be divisible by a number M (with M non-zero) if, when we divide N by M, the remainder of the division is zero.

$N mod M = 0$

Brute force algorithm

We could start by using a brute-force method which would involve dividing 9 by each of the numbers from 1 to 9 to determine which ones have a remainder equal to 0.

$Remainder = N - (M \times ⌊ \frac{N}{M} ⌋)$

(where $⌊ \frac{N}{M} ⌋$ is the integer part of the quotient)

9 / 1 = 9 (the remainder is 0, so 1 is a divisor of 9)
9 / 2 = 4.5 (the remainder is 1, so 2 is not a divisor of 9)
9 / 3 = 3 (the remainder is 0, so 3 is a divisor of 9)
...
9 / 4 = 2.25 (the remainder is 1, so 4 is not a divisor of 9)
9 / 5 = 1.8 (the remainder is 4, so 5 is not a divisor of 9)
9 / 6 = 1.5 (the remainder is 3, so 6 is not a divisor of 9)
9 / 7 = 1.2857142857143 (the remainder is 2, so 7 is not a divisor of 9)
9 / 8 = 1.125 (the remainder is 1, so 8 is not a divisor of 9)
9 / 9 = 1 (the remainder is 0, so 9 is a divisor of 9)

Improved algorithm using square-root

However, there is another slightly better approach that reduces the number of iterations by testing only integers less than or equal to the square root of 9 (i.e. 3). Indeed, if a number N has a divisor D greater than its square root, then there is necessarily a smaller divisor d such that:

$D \times d = N$

(thus, if $\frac{N}{D} = d$ , then $\frac{N}{d} = D$ )

9 / 1 = 9 (the remainder is 0, so 1 and 9 are divisors of 9)
9 / 2 = 4.5 (the remainder is 1, so 2 is not a divisor of 9)
9 / 3 = 3 (the remainder is 0, so 3 and 3 are divisors of 9)