What are the divisors of 5733?

1, 3, 7, 9, 13, 21, 39, 49, 63, 91, 117, 147, 273, 441, 637, 819, 1911, 5733

There is a total of 18 positive divisors.
The sum of these divisors is 10374.
The arithmetic mean is 576.33333333333.

18 odd divisors

1, 3, 7, 9, 13, 21, 39, 49, 63, 91, 117, 147, 273, 441, 637, 819, 1911, 5733

How to compute the divisors of 5733?

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 5733 by each of the numbers from 1 to 5733 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)

5733 / 1 = 5733 (the remainder is 0, so 1 is a divisor of 5733)
5733 / 2 = 2866.5 (the remainder is 1, so 2 is not a divisor of 5733)
5733 / 3 = 1911 (the remainder is 0, so 3 is a divisor of 5733)
...
5733 / 5732 = 1.0001744591766 (the remainder is 1, so 5732 is not a divisor of 5733)
5733 / 5733 = 1 (the remainder is 0, so 5733 is a divisor of 5733)

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 5733 (i.e. 75.716576784744). 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 $\frac{N}{d} = D$ )

5733 / 1 = 5733 (the remainder is 0, so 1 and 5733 are divisors of 5733)
5733 / 2 = 2866.5 (the remainder is 1, so 2 is not a divisor of 5733)
5733 / 3 = 1911 (the remainder is 0, so 3 and 1911 are divisors of 5733)
...
5733 / 74 = 77.472972972973 (the remainder is 35, so 74 is not a divisor of 5733)
5733 / 75 = 76.44 (the remainder is 33, so 75 is not a divisor of 5733)