What are the divisors of 5823?

1, 3, 9, 647, 1941, 5823

There is a total of 6 positive divisors.
The sum of these divisors is 8424.
The arithmetic mean is 1404.

6 odd divisors

1, 3, 9, 647, 1941, 5823

How to compute the divisors of 5823?

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 5823 by each of the numbers from 1 to 5823 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)

5823 / 1 = 5823 (the remainder is 0, so 1 is a divisor of 5823)
5823 / 2 = 2911.5 (the remainder is 1, so 2 is not a divisor of 5823)
5823 / 3 = 1941 (the remainder is 0, so 3 is a divisor of 5823)
...
5823 / 5822 = 1.000171762281 (the remainder is 1, so 5822 is not a divisor of 5823)
5823 / 5823 = 1 (the remainder is 0, so 5823 is a divisor of 5823)

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 5823 (i.e. 76.308584051861). 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 $\frac{N}{d} = D$ )

5823 / 1 = 5823 (the remainder is 0, so 1 and 5823 are divisors of 5823)
5823 / 2 = 2911.5 (the remainder is 1, so 2 is not a divisor of 5823)
5823 / 3 = 1941 (the remainder is 0, so 3 and 1941 are divisors of 5823)
...
5823 / 75 = 77.64 (the remainder is 48, so 75 is not a divisor of 5823)
5823 / 76 = 76.618421052632 (the remainder is 47, so 76 is not a divisor of 5823)