What are the divisors of 5803?

1, 7, 829, 5803

There is a total of 4 positive divisors.
The sum of these divisors is 6640.
The arithmetic mean is 1660.

4 odd divisors

1, 7, 829, 5803

How to compute the divisors of 5803?

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 5803 by each of the numbers from 1 to 5803 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)

5803 / 1 = 5803 (the remainder is 0, so 1 is a divisor of 5803)
5803 / 2 = 2901.5 (the remainder is 1, so 2 is not a divisor of 5803)
5803 / 3 = 1934.3333333333 (the remainder is 1, so 3 is not a divisor of 5803)
...
5803 / 5802 = 1.0001723543606 (the remainder is 1, so 5802 is not a divisor of 5803)
5803 / 5803 = 1 (the remainder is 0, so 5803 is a divisor of 5803)

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 5803 (i.e. 76.177424477334). 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 $\frac{N}{d} = D$ )

5803 / 1 = 5803 (the remainder is 0, so 1 and 5803 are divisors of 5803)
5803 / 2 = 2901.5 (the remainder is 1, so 2 is not a divisor of 5803)
5803 / 3 = 1934.3333333333 (the remainder is 1, so 3 is not a divisor of 5803)
...
5803 / 75 = 77.373333333333 (the remainder is 28, so 75 is not a divisor of 5803)
5803 / 76 = 76.355263157895 (the remainder is 27, so 76 is not a divisor of 5803)