What are the divisors of 8012?

1, 2, 4, 2003, 4006, 8012

There is a total of 6 positive divisors.
The sum of these divisors is 14028.
The arithmetic mean is 2338.

4 even divisors

2, 4, 4006, 8012

2 odd divisors

1, 2003

How to compute the divisors of 8012?

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 8012 by each of the numbers from 1 to 8012 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)

8012 / 1 = 8012 (the remainder is 0, so 1 is a divisor of 8012)
8012 / 2 = 4006 (the remainder is 0, so 2 is a divisor of 8012)
8012 / 3 = 2670.6666666667 (the remainder is 2, so 3 is not a divisor of 8012)
...
8012 / 8011 = 1.000124828361 (the remainder is 1, so 8011 is not a divisor of 8012)
8012 / 8012 = 1 (the remainder is 0, so 8012 is a divisor of 8012)

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 8012 (i.e. 89.509776002401). 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 $\frac{N}{d} = D$ )

8012 / 1 = 8012 (the remainder is 0, so 1 and 8012 are divisors of 8012)
8012 / 2 = 4006 (the remainder is 0, so 2 and 4006 are divisors of 8012)
8012 / 3 = 2670.6666666667 (the remainder is 2, so 3 is not a divisor of 8012)
...
8012 / 88 = 91.045454545455 (the remainder is 4, so 88 is not a divisor of 8012)
8012 / 89 = 90.022471910112 (the remainder is 2, so 89 is not a divisor of 8012)