What are the divisors of 5703?

1, 3, 1901, 5703

There is a total of 4 positive divisors.
The sum of these divisors is 7608.
The arithmetic mean is 1902.

4 odd divisors

1, 3, 1901, 5703

How to compute the divisors of 5703?

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 5703 by each of the numbers from 1 to 5703 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)

5703 / 1 = 5703 (the remainder is 0, so 1 is a divisor of 5703)
5703 / 2 = 2851.5 (the remainder is 1, so 2 is not a divisor of 5703)
5703 / 3 = 1901 (the remainder is 0, so 3 is a divisor of 5703)
...
5703 / 5702 = 1.0001753770607 (the remainder is 1, so 5702 is not a divisor of 5703)
5703 / 5703 = 1 (the remainder is 0, so 5703 is a divisor of 5703)

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 5703 (i.e. 75.518209724543). 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 $\frac{N}{d} = D$ )

5703 / 1 = 5703 (the remainder is 0, so 1 and 5703 are divisors of 5703)
5703 / 2 = 2851.5 (the remainder is 1, so 2 is not a divisor of 5703)
5703 / 3 = 1901 (the remainder is 0, so 3 and 1901 are divisors of 5703)
...
5703 / 74 = 77.067567567568 (the remainder is 5, so 74 is not a divisor of 5703)
5703 / 75 = 76.04 (the remainder is 3, so 75 is not a divisor of 5703)