What are the divisors of 5238?

1, 2, 3, 6, 9, 18, 27, 54, 97, 194, 291, 582, 873, 1746, 2619, 5238

There is a total of 16 positive divisors.
The sum of these divisors is 11760.
The arithmetic mean is 735.

8 even divisors

2, 6, 18, 54, 194, 582, 1746, 5238

8 odd divisors

1, 3, 9, 27, 97, 291, 873, 2619

How to compute the divisors of 5238?

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 5238 by each of the numbers from 1 to 5238 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)

5238 / 1 = 5238 (the remainder is 0, so 1 is a divisor of 5238)
5238 / 2 = 2619 (the remainder is 0, so 2 is a divisor of 5238)
5238 / 3 = 1746 (the remainder is 0, so 3 is a divisor of 5238)
...
5238 / 5237 = 1.0001909490166 (the remainder is 1, so 5237 is not a divisor of 5238)
5238 / 5238 = 1 (the remainder is 0, so 5238 is a divisor of 5238)

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 5238 (i.e. 72.374028490889). 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 $\frac{N}{d} = D$ )

5238 / 1 = 5238 (the remainder is 0, so 1 and 5238 are divisors of 5238)
5238 / 2 = 2619 (the remainder is 0, so 2 and 2619 are divisors of 5238)
5238 / 3 = 1746 (the remainder is 0, so 3 and 1746 are divisors of 5238)
...
5238 / 71 = 73.774647887324 (the remainder is 55, so 71 is not a divisor of 5238)
5238 / 72 = 72.75 (the remainder is 54, so 72 is not a divisor of 5238)