What are the divisors of 5382?

1, 2, 3, 6, 9, 13, 18, 23, 26, 39, 46, 69, 78, 117, 138, 207, 234, 299, 414, 598, 897, 1794, 2691, 5382

There is a total of 24 positive divisors.
The sum of these divisors is 13104.
The arithmetic mean is 546.

12 even divisors

2, 6, 18, 26, 46, 78, 138, 234, 414, 598, 1794, 5382

12 odd divisors

1, 3, 9, 13, 23, 39, 69, 117, 207, 299, 897, 2691

How to compute the divisors of 5382?

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 5382 by each of the numbers from 1 to 5382 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)

5382 / 1 = 5382 (the remainder is 0, so 1 is a divisor of 5382)
5382 / 2 = 2691 (the remainder is 0, so 2 is a divisor of 5382)
5382 / 3 = 1794 (the remainder is 0, so 3 is a divisor of 5382)
...
5382 / 5381 = 1.0001858390634 (the remainder is 1, so 5381 is not a divisor of 5382)
5382 / 5382 = 1 (the remainder is 0, so 5382 is a divisor of 5382)

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 5382 (i.e. 73.362115563825). 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 $\frac{N}{d} = D$ )

5382 / 1 = 5382 (the remainder is 0, so 1 and 5382 are divisors of 5382)
5382 / 2 = 2691 (the remainder is 0, so 2 and 2691 are divisors of 5382)
5382 / 3 = 1794 (the remainder is 0, so 3 and 1794 are divisors of 5382)
...
5382 / 72 = 74.75 (the remainder is 54, so 72 is not a divisor of 5382)
5382 / 73 = 73.72602739726 (the remainder is 53, so 73 is not a divisor of 5382)