What are the divisors of 9090?

1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90, 101, 202, 303, 505, 606, 909, 1010, 1515, 1818, 3030, 4545, 9090

There is a total of 24 positive divisors.
The sum of these divisors is 23868.
The arithmetic mean is 994.5.

12 even divisors

2, 6, 10, 18, 30, 90, 202, 606, 1010, 1818, 3030, 9090

12 odd divisors

1, 3, 5, 9, 15, 45, 101, 303, 505, 909, 1515, 4545

How to compute the divisors of 9090?

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 9090 by each of the numbers from 1 to 9090 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)

9090 / 1 = 9090 (the remainder is 0, so 1 is a divisor of 9090)
9090 / 2 = 4545 (the remainder is 0, so 2 is a divisor of 9090)
9090 / 3 = 3030 (the remainder is 0, so 3 is a divisor of 9090)
...
9090 / 9089 = 1.0001100231049 (the remainder is 1, so 9089 is not a divisor of 9090)
9090 / 9090 = 1 (the remainder is 0, so 9090 is a divisor of 9090)

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 9090 (i.e. 95.341491492424). 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 $\frac{N}{d} = D$ )

9090 / 1 = 9090 (the remainder is 0, so 1 and 9090 are divisors of 9090)
9090 / 2 = 4545 (the remainder is 0, so 2 and 4545 are divisors of 9090)
9090 / 3 = 3030 (the remainder is 0, so 3 and 3030 are divisors of 9090)
...
9090 / 94 = 96.702127659574 (the remainder is 66, so 94 is not a divisor of 9090)
9090 / 95 = 95.684210526316 (the remainder is 65, so 95 is not a divisor of 9090)