What are the divisors of 8268?

1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 53, 78, 106, 156, 159, 212, 318, 636, 689, 1378, 2067, 2756, 4134, 8268

There is a total of 24 positive divisors.
The sum of these divisors is 21168.
The arithmetic mean is 882.

16 even divisors

2, 4, 6, 12, 26, 52, 78, 106, 156, 212, 318, 636, 1378, 2756, 4134, 8268

8 odd divisors

1, 3, 13, 39, 53, 159, 689, 2067

How to compute the divisors of 8268?

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 8268 by each of the numbers from 1 to 8268 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)

8268 / 1 = 8268 (the remainder is 0, so 1 is a divisor of 8268)
8268 / 2 = 4134 (the remainder is 0, so 2 is a divisor of 8268)
8268 / 3 = 2756 (the remainder is 0, so 3 is a divisor of 8268)
...
8268 / 8267 = 1.0001209628644 (the remainder is 1, so 8267 is not a divisor of 8268)
8268 / 8268 = 1 (the remainder is 0, so 8268 is a divisor of 8268)

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 8268 (i.e. 90.928543373354). 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 $\frac{N}{d} = D$ )

8268 / 1 = 8268 (the remainder is 0, so 1 and 8268 are divisors of 8268)
8268 / 2 = 4134 (the remainder is 0, so 2 and 4134 are divisors of 8268)
8268 / 3 = 2756 (the remainder is 0, so 3 and 2756 are divisors of 8268)
...
8268 / 89 = 92.898876404494 (the remainder is 80, so 89 is not a divisor of 8268)
8268 / 90 = 91.866666666667 (the remainder is 78, so 90 is not a divisor of 8268)