What are the divisors of 2805?

1, 3, 5, 11, 15, 17, 33, 51, 55, 85, 165, 187, 255, 561, 935, 2805

There is a total of 16 positive divisors.
The sum of these divisors is 5184.
The arithmetic mean is 324.

16 odd divisors

1, 3, 5, 11, 15, 17, 33, 51, 55, 85, 165, 187, 255, 561, 935, 2805

How to compute the divisors of 2805?

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 2805 by each of the numbers from 1 to 2805 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)

2805 / 1 = 2805 (the remainder is 0, so 1 is a divisor of 2805)
2805 / 2 = 1402.5 (the remainder is 1, so 2 is not a divisor of 2805)
2805 / 3 = 935 (the remainder is 0, so 3 is a divisor of 2805)
...
2805 / 2804 = 1.0003566333809 (the remainder is 1, so 2804 is not a divisor of 2805)
2805 / 2805 = 1 (the remainder is 0, so 2805 is a divisor of 2805)

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 2805 (i.e. 52.962250707461). 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 $\frac{N}{d} = D$ )

2805 / 1 = 2805 (the remainder is 0, so 1 and 2805 are divisors of 2805)
2805 / 2 = 1402.5 (the remainder is 1, so 2 is not a divisor of 2805)
2805 / 3 = 935 (the remainder is 0, so 3 and 935 are divisors of 2805)
...
2805 / 51 = 55 (the remainder is 0, so 51 and 55 are divisors of 2805)
2805 / 52 = 53.942307692308 (the remainder is 49, so 52 is not a divisor of 2805)