What are the divisors of 9150?

1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 61, 75, 122, 150, 183, 305, 366, 610, 915, 1525, 1830, 3050, 4575, 9150

There is a total of 24 positive divisors.
The sum of these divisors is 23064.
The arithmetic mean is 961.

12 even divisors

2, 6, 10, 30, 50, 122, 150, 366, 610, 1830, 3050, 9150

12 odd divisors

1, 3, 5, 15, 25, 61, 75, 183, 305, 915, 1525, 4575

How to compute the divisors of 9150?

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 9150 by each of the numbers from 1 to 9150 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)

9150 / 1 = 9150 (the remainder is 0, so 1 is a divisor of 9150)
9150 / 2 = 4575 (the remainder is 0, so 2 is a divisor of 9150)
9150 / 3 = 3050 (the remainder is 0, so 3 is a divisor of 9150)
...
9150 / 9149 = 1.000109301563 (the remainder is 1, so 9149 is not a divisor of 9150)
9150 / 9150 = 1 (the remainder is 0, so 9150 is a divisor of 9150)

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 9150 (i.e. 95.655632348545). 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 $\frac{N}{d} = D$ )

9150 / 1 = 9150 (the remainder is 0, so 1 and 9150 are divisors of 9150)
9150 / 2 = 4575 (the remainder is 0, so 2 and 4575 are divisors of 9150)
9150 / 3 = 3050 (the remainder is 0, so 3 and 3050 are divisors of 9150)
...
9150 / 94 = 97.340425531915 (the remainder is 32, so 94 is not a divisor of 9150)
9150 / 95 = 96.315789473684 (the remainder is 30, so 95 is not a divisor of 9150)