What are the divisors of 990?

1, 2, 3, 5, 6, 9, 10, 11, 15, 18, 22, 30, 33, 45, 55, 66, 90, 99, 110, 165, 198, 330, 495, 990

There is a total of 24 positive divisors.
The sum of these divisors is 2808.
The arithmetic mean is 117.

12 even divisors

2, 6, 10, 18, 22, 30, 66, 90, 110, 198, 330, 990

12 odd divisors

1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165, 495

How to compute the divisors of 990?

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 990 by each of the numbers from 1 to 990 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)

990 / 1 = 990 (the remainder is 0, so 1 is a divisor of 990)
990 / 2 = 495 (the remainder is 0, so 2 is a divisor of 990)
990 / 3 = 330 (the remainder is 0, so 3 is a divisor of 990)
...
990 / 989 = 1.0010111223458 (the remainder is 1, so 989 is not a divisor of 990)
990 / 990 = 1 (the remainder is 0, so 990 is a divisor of 990)

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 990 (i.e. 31.464265445105). 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 $\frac{N}{d} = D$ )

990 / 1 = 990 (the remainder is 0, so 1 and 990 are divisors of 990)
990 / 2 = 495 (the remainder is 0, so 2 and 495 are divisors of 990)
990 / 3 = 330 (the remainder is 0, so 3 and 330 are divisors of 990)
...
990 / 30 = 33 (the remainder is 0, so 30 and 33 are divisors of 990)
990 / 31 = 31.935483870968 (the remainder is 29, so 31 is not a divisor of 990)