What are the divisors of 8037?

1, 3, 9, 19, 47, 57, 141, 171, 423, 893, 2679, 8037

There is a total of 12 positive divisors.
The sum of these divisors is 12480.
The arithmetic mean is 1040.

12 odd divisors

1, 3, 9, 19, 47, 57, 141, 171, 423, 893, 2679, 8037

How to compute the divisors of 8037?

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 8037 by each of the numbers from 1 to 8037 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)

8037 / 1 = 8037 (the remainder is 0, so 1 is a divisor of 8037)
8037 / 2 = 4018.5 (the remainder is 1, so 2 is not a divisor of 8037)
8037 / 3 = 2679 (the remainder is 0, so 3 is a divisor of 8037)
...
8037 / 8036 = 1.0001244400199 (the remainder is 1, so 8036 is not a divisor of 8037)
8037 / 8037 = 1 (the remainder is 0, so 8037 is a divisor of 8037)

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 8037 (i.e. 89.649316784904). 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 $\frac{N}{d} = D$ )

8037 / 1 = 8037 (the remainder is 0, so 1 and 8037 are divisors of 8037)
8037 / 2 = 4018.5 (the remainder is 1, so 2 is not a divisor of 8037)
8037 / 3 = 2679 (the remainder is 0, so 3 and 2679 are divisors of 8037)
...
8037 / 88 = 91.329545454545 (the remainder is 29, so 88 is not a divisor of 8037)
8037 / 89 = 90.303370786517 (the remainder is 27, so 89 is not a divisor of 8037)