What are the divisors of 8109?

1, 3, 9, 17, 51, 53, 153, 159, 477, 901, 2703, 8109

There is a total of 12 positive divisors.
The sum of these divisors is 12636.
The arithmetic mean is 1053.

12 odd divisors

1, 3, 9, 17, 51, 53, 153, 159, 477, 901, 2703, 8109

How to compute the divisors of 8109?

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 8109 by each of the numbers from 1 to 8109 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)

8109 / 1 = 8109 (the remainder is 0, so 1 is a divisor of 8109)
8109 / 2 = 4054.5 (the remainder is 1, so 2 is not a divisor of 8109)
8109 / 3 = 2703 (the remainder is 0, so 3 is a divisor of 8109)
...
8109 / 8108 = 1.0001233349778 (the remainder is 1, so 8108 is not a divisor of 8109)
8109 / 8109 = 1 (the remainder is 0, so 8109 is a divisor of 8109)

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 8109 (i.e. 90.049986118822). 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 $\frac{N}{d} = D$ )

8109 / 1 = 8109 (the remainder is 0, so 1 and 8109 are divisors of 8109)
8109 / 2 = 4054.5 (the remainder is 1, so 2 is not a divisor of 8109)
8109 / 3 = 2703 (the remainder is 0, so 3 and 2703 are divisors of 8109)
...
8109 / 89 = 91.112359550562 (the remainder is 10, so 89 is not a divisor of 8109)
8109 / 90 = 90.1 (the remainder is 9, so 90 is not a divisor of 8109)