What are the divisors of 8010?

1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 89, 90, 178, 267, 445, 534, 801, 890, 1335, 1602, 2670, 4005, 8010

There is a total of 24 positive divisors.
The sum of these divisors is 21060.
The arithmetic mean is 877.5.

12 even divisors

2, 6, 10, 18, 30, 90, 178, 534, 890, 1602, 2670, 8010

12 odd divisors

1, 3, 5, 9, 15, 45, 89, 267, 445, 801, 1335, 4005

How to compute the divisors of 8010?

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 8010 by each of the numbers from 1 to 8010 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)

8010 / 1 = 8010 (the remainder is 0, so 1 is a divisor of 8010)
8010 / 2 = 4005 (the remainder is 0, so 2 is a divisor of 8010)
8010 / 3 = 2670 (the remainder is 0, so 3 is a divisor of 8010)
...
8010 / 8009 = 1.000124859533 (the remainder is 1, so 8009 is not a divisor of 8010)
8010 / 8010 = 1 (the remainder is 0, so 8010 is a divisor of 8010)

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 8010 (i.e. 89.498603341058). 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 $\frac{N}{d} = D$ )

8010 / 1 = 8010 (the remainder is 0, so 1 and 8010 are divisors of 8010)
8010 / 2 = 4005 (the remainder is 0, so 2 and 4005 are divisors of 8010)
8010 / 3 = 2670 (the remainder is 0, so 3 and 2670 are divisors of 8010)
...
8010 / 88 = 91.022727272727 (the remainder is 2, so 88 is not a divisor of 8010)
8010 / 89 = 90 (the remainder is 0, so 89 and 90 are divisors of 8010)