What are the divisors of 2205?

1, 3, 5, 7, 9, 15, 21, 35, 45, 49, 63, 105, 147, 245, 315, 441, 735, 2205

There is a total of 18 positive divisors.
The sum of these divisors is 4446.
The arithmetic mean is 247.

18 odd divisors

1, 3, 5, 7, 9, 15, 21, 35, 45, 49, 63, 105, 147, 245, 315, 441, 735, 2205

How to compute the divisors of 2205?

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 2205 by each of the numbers from 1 to 2205 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)

2205 / 1 = 2205 (the remainder is 0, so 1 is a divisor of 2205)
2205 / 2 = 1102.5 (the remainder is 1, so 2 is not a divisor of 2205)
2205 / 3 = 735 (the remainder is 0, so 3 is a divisor of 2205)
...
2205 / 2204 = 1.0004537205082 (the remainder is 1, so 2204 is not a divisor of 2205)
2205 / 2205 = 1 (the remainder is 0, so 2205 is a divisor of 2205)

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 2205 (i.e. 46.957427527496). 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 $\frac{N}{d} = D$ )

2205 / 1 = 2205 (the remainder is 0, so 1 and 2205 are divisors of 2205)
2205 / 2 = 1102.5 (the remainder is 1, so 2 is not a divisor of 2205)
2205 / 3 = 735 (the remainder is 0, so 3 and 735 are divisors of 2205)
...
2205 / 45 = 49 (the remainder is 0, so 45 and 49 are divisors of 2205)
2205 / 46 = 47.934782608696 (the remainder is 43, so 46 is not a divisor of 2205)