What are the divisors of 8038?

1, 2, 4019, 8038

There is a total of 4 positive divisors.
The sum of these divisors is 12060.
The arithmetic mean is 3015.

2 even divisors

2, 8038

2 odd divisors

1, 4019

How to compute the divisors of 8038?

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 8038 by each of the numbers from 1 to 8038 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)

8038 / 1 = 8038 (the remainder is 0, so 1 is a divisor of 8038)
8038 / 2 = 4019 (the remainder is 0, so 2 is a divisor of 8038)
8038 / 3 = 2679.3333333333 (the remainder is 1, so 3 is not a divisor of 8038)
...
8038 / 8037 = 1.0001244245365 (the remainder is 1, so 8037 is not a divisor of 8038)
8038 / 8038 = 1 (the remainder is 0, so 8038 is a divisor of 8038)

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 8038 (i.e. 89.654893898772). 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 $\frac{N}{d} = D$ )

8038 / 1 = 8038 (the remainder is 0, so 1 and 8038 are divisors of 8038)
8038 / 2 = 4019 (the remainder is 0, so 2 and 4019 are divisors of 8038)
8038 / 3 = 2679.3333333333 (the remainder is 1, so 3 is not a divisor of 8038)
...
8038 / 88 = 91.340909090909 (the remainder is 30, so 88 is not a divisor of 8038)
8038 / 89 = 90.314606741573 (the remainder is 28, so 89 is not a divisor of 8038)