What are the divisors of 8055?

1, 3, 5, 9, 15, 45, 179, 537, 895, 1611, 2685, 8055

There is a total of 12 positive divisors.
The sum of these divisors is 14040.
The arithmetic mean is 1170.

12 odd divisors

1, 3, 5, 9, 15, 45, 179, 537, 895, 1611, 2685, 8055

How to compute the divisors of 8055?

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 8055 by each of the numbers from 1 to 8055 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)

8055 / 1 = 8055 (the remainder is 0, so 1 is a divisor of 8055)
8055 / 2 = 4027.5 (the remainder is 1, so 2 is not a divisor of 8055)
8055 / 3 = 2685 (the remainder is 0, so 3 is a divisor of 8055)
...
8055 / 8054 = 1.0001241619071 (the remainder is 1, so 8054 is not a divisor of 8055)
8055 / 8055 = 1 (the remainder is 0, so 8055 is a divisor of 8055)

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 8055 (i.e. 89.74965180991). 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 $\frac{N}{d} = D$ )

8055 / 1 = 8055 (the remainder is 0, so 1 and 8055 are divisors of 8055)
8055 / 2 = 4027.5 (the remainder is 1, so 2 is not a divisor of 8055)
8055 / 3 = 2685 (the remainder is 0, so 3 and 2685 are divisors of 8055)
...
8055 / 88 = 91.534090909091 (the remainder is 47, so 88 is not a divisor of 8055)
8055 / 89 = 90.505617977528 (the remainder is 45, so 89 is not a divisor of 8055)