What are the divisors of 5806?

1, 2, 2903, 5806

There is a total of 4 positive divisors.
The sum of these divisors is 8712.
The arithmetic mean is 2178.

2 even divisors

2, 5806

2 odd divisors

1, 2903

How to compute the divisors of 5806?

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 5806 by each of the numbers from 1 to 5806 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)

5806 / 1 = 5806 (the remainder is 0, so 1 is a divisor of 5806)
5806 / 2 = 2903 (the remainder is 0, so 2 is a divisor of 5806)
5806 / 3 = 1935.3333333333 (the remainder is 1, so 3 is not a divisor of 5806)
...
5806 / 5805 = 1.0001722652885 (the remainder is 1, so 5805 is not a divisor of 5806)
5806 / 5806 = 1 (the remainder is 0, so 5806 is a divisor of 5806)

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 5806 (i.e. 76.197112806195). 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 $\frac{N}{d} = D$ )

5806 / 1 = 5806 (the remainder is 0, so 1 and 5806 are divisors of 5806)
5806 / 2 = 2903 (the remainder is 0, so 2 and 2903 are divisors of 5806)
5806 / 3 = 1935.3333333333 (the remainder is 1, so 3 is not a divisor of 5806)
...
5806 / 75 = 77.413333333333 (the remainder is 31, so 75 is not a divisor of 5806)
5806 / 76 = 76.394736842105 (the remainder is 30, so 76 is not a divisor of 5806)