What are the divisors of 2004?

1, 2, 3, 4, 6, 12, 167, 334, 501, 668, 1002, 2004

There is a total of 12 positive divisors.
The sum of these divisors is 4704.
The arithmetic mean is 392.

8 even divisors

2, 4, 6, 12, 334, 668, 1002, 2004

4 odd divisors

1, 3, 167, 501

How to compute the divisors of 2004?

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 2004 by each of the numbers from 1 to 2004 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)

2004 / 1 = 2004 (the remainder is 0, so 1 is a divisor of 2004)
2004 / 2 = 1002 (the remainder is 0, so 2 is a divisor of 2004)
2004 / 3 = 668 (the remainder is 0, so 3 is a divisor of 2004)
...
2004 / 2003 = 1.0004992511233 (the remainder is 1, so 2003 is not a divisor of 2004)
2004 / 2004 = 1 (the remainder is 0, so 2004 is a divisor of 2004)

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 2004 (i.e. 44.766058571199). 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 $\frac{N}{d} = D$ )

2004 / 1 = 2004 (the remainder is 0, so 1 and 2004 are divisors of 2004)
2004 / 2 = 1002 (the remainder is 0, so 2 and 1002 are divisors of 2004)
2004 / 3 = 668 (the remainder is 0, so 3 and 668 are divisors of 2004)
...
2004 / 43 = 46.604651162791 (the remainder is 26, so 43 is not a divisor of 2004)
2004 / 44 = 45.545454545455 (the remainder is 24, so 44 is not a divisor of 2004)