What are the divisors of 8070?

1, 2, 3, 5, 6, 10, 15, 30, 269, 538, 807, 1345, 1614, 2690, 4035, 8070

There is a total of 16 positive divisors.
The sum of these divisors is 19440.
The arithmetic mean is 1215.

8 even divisors

2, 6, 10, 30, 538, 1614, 2690, 8070

8 odd divisors

1, 3, 5, 15, 269, 807, 1345, 4035

How to compute the divisors of 8070?

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 8070 by each of the numbers from 1 to 8070 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)

8070 / 1 = 8070 (the remainder is 0, so 1 is a divisor of 8070)
8070 / 2 = 4035 (the remainder is 0, so 2 is a divisor of 8070)
8070 / 3 = 2690 (the remainder is 0, so 3 is a divisor of 8070)
...
8070 / 8069 = 1.0001239310943 (the remainder is 1, so 8069 is not a divisor of 8070)
8070 / 8070 = 1 (the remainder is 0, so 8070 is a divisor of 8070)

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 8070 (i.e. 89.833178725903). 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 $\frac{N}{d} = D$ )

8070 / 1 = 8070 (the remainder is 0, so 1 and 8070 are divisors of 8070)
8070 / 2 = 4035 (the remainder is 0, so 2 and 4035 are divisors of 8070)
8070 / 3 = 2690 (the remainder is 0, so 3 and 2690 are divisors of 8070)
...
8070 / 88 = 91.704545454545 (the remainder is 62, so 88 is not a divisor of 8070)
8070 / 89 = 90.674157303371 (the remainder is 60, so 89 is not a divisor of 8070)