What are the divisors of 4008?

1, 2, 3, 4, 6, 8, 12, 24, 167, 334, 501, 668, 1002, 1336, 2004, 4008

There is a total of 16 positive divisors.
The sum of these divisors is 10080.
The arithmetic mean is 630.

12 even divisors

2, 4, 6, 8, 12, 24, 334, 668, 1002, 1336, 2004, 4008

4 odd divisors

1, 3, 167, 501

How to compute the divisors of 4008?

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 4008 by each of the numbers from 1 to 4008 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)

4008 / 1 = 4008 (the remainder is 0, so 1 is a divisor of 4008)
4008 / 2 = 2004 (the remainder is 0, so 2 is a divisor of 4008)
4008 / 3 = 1336 (the remainder is 0, so 3 is a divisor of 4008)
...
4008 / 4007 = 1.0002495632643 (the remainder is 1, so 4007 is not a divisor of 4008)
4008 / 4008 = 1 (the remainder is 0, so 4008 is a divisor of 4008)

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 4008 (i.e. 63.308767165378). 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 $\frac{N}{d} = D$ )

4008 / 1 = 4008 (the remainder is 0, so 1 and 4008 are divisors of 4008)
4008 / 2 = 2004 (the remainder is 0, so 2 and 2004 are divisors of 4008)
4008 / 3 = 1336 (the remainder is 0, so 3 and 1336 are divisors of 4008)
...
4008 / 62 = 64.645161290323 (the remainder is 40, so 62 is not a divisor of 4008)
4008 / 63 = 63.619047619048 (the remainder is 39, so 63 is not a divisor of 4008)