What are the divisors of 9135?

1, 3, 5, 7, 9, 15, 21, 29, 35, 45, 63, 87, 105, 145, 203, 261, 315, 435, 609, 1015, 1305, 1827, 3045, 9135

There is a total of 24 positive divisors.
The sum of these divisors is 18720.
The arithmetic mean is 780.

24 odd divisors

1, 3, 5, 7, 9, 15, 21, 29, 35, 45, 63, 87, 105, 145, 203, 261, 315, 435, 609, 1015, 1305, 1827, 3045, 9135

How to compute the divisors of 9135?

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 9135 by each of the numbers from 1 to 9135 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)

9135 / 1 = 9135 (the remainder is 0, so 1 is a divisor of 9135)
9135 / 2 = 4567.5 (the remainder is 1, so 2 is not a divisor of 9135)
9135 / 3 = 3045 (the remainder is 0, so 3 is a divisor of 9135)
...
9135 / 9134 = 1.0001094810598 (the remainder is 1, so 9134 is not a divisor of 9135)
9135 / 9135 = 1 (the remainder is 0, so 9135 is a divisor of 9135)

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 9135 (i.e. 95.577193932444). 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 $\frac{N}{d} = D$ )

9135 / 1 = 9135 (the remainder is 0, so 1 and 9135 are divisors of 9135)
9135 / 2 = 4567.5 (the remainder is 1, so 2 is not a divisor of 9135)
9135 / 3 = 3045 (the remainder is 0, so 3 and 3045 are divisors of 9135)
...
9135 / 94 = 97.18085106383 (the remainder is 17, so 94 is not a divisor of 9135)
9135 / 95 = 96.157894736842 (the remainder is 15, so 95 is not a divisor of 9135)