What are the divisors of 2005?

1, 5, 401, 2005

There is a total of 4 positive divisors.
The sum of these divisors is 2412.
The arithmetic mean is 603.

4 odd divisors

1, 5, 401, 2005

How to compute the divisors of 2005?

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 2005 by each of the numbers from 1 to 2005 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)

2005 / 1 = 2005 (the remainder is 0, so 1 is a divisor of 2005)
2005 / 2 = 1002.5 (the remainder is 1, so 2 is not a divisor of 2005)
2005 / 3 = 668.33333333333 (the remainder is 1, so 3 is not a divisor of 2005)
...
2005 / 2004 = 1.000499001996 (the remainder is 1, so 2004 is not a divisor of 2005)
2005 / 2005 = 1 (the remainder is 0, so 2005 is a divisor of 2005)

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 2005 (i.e. 44.777226354476). 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 $\frac{N}{d} = D$ )

2005 / 1 = 2005 (the remainder is 0, so 1 and 2005 are divisors of 2005)
2005 / 2 = 1002.5 (the remainder is 1, so 2 is not a divisor of 2005)
2005 / 3 = 668.33333333333 (the remainder is 1, so 3 is not a divisor of 2005)
...
2005 / 43 = 46.627906976744 (the remainder is 27, so 43 is not a divisor of 2005)
2005 / 44 = 45.568181818182 (the remainder is 25, so 44 is not a divisor of 2005)