What are the divisors of 1815?

1, 3, 5, 11, 15, 33, 55, 121, 165, 363, 605, 1815

There is a total of 12 positive divisors.
The sum of these divisors is 3192.
The arithmetic mean is 266.

12 odd divisors

1, 3, 5, 11, 15, 33, 55, 121, 165, 363, 605, 1815

How to compute the divisors of 1815?

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 1815 by each of the numbers from 1 to 1815 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)

1815 / 1 = 1815 (the remainder is 0, so 1 is a divisor of 1815)
1815 / 2 = 907.5 (the remainder is 1, so 2 is not a divisor of 1815)
1815 / 3 = 605 (the remainder is 0, so 3 is a divisor of 1815)
...
1815 / 1814 = 1.0005512679162 (the remainder is 1, so 1814 is not a divisor of 1815)
1815 / 1815 = 1 (the remainder is 0, so 1815 is a divisor of 1815)

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 1815 (i.e. 42.602816808282). 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 $\frac{N}{d} = D$ )

1815 / 1 = 1815 (the remainder is 0, so 1 and 1815 are divisors of 1815)
1815 / 2 = 907.5 (the remainder is 1, so 2 is not a divisor of 1815)
1815 / 3 = 605 (the remainder is 0, so 3 and 605 are divisors of 1815)
...
1815 / 41 = 44.268292682927 (the remainder is 11, so 41 is not a divisor of 1815)
1815 / 42 = 43.214285714286 (the remainder is 9, so 42 is not a divisor of 1815)