What are the divisors of 5438?

1, 2, 2719, 5438

There is a total of 4 positive divisors.
The sum of these divisors is 8160.
The arithmetic mean is 2040.

2 even divisors

2, 5438

2 odd divisors

1, 2719

How to compute the divisors of 5438?

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 5438 by each of the numbers from 1 to 5438 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)

5438 / 1 = 5438 (the remainder is 0, so 1 is a divisor of 5438)
5438 / 2 = 2719 (the remainder is 0, so 2 is a divisor of 5438)
5438 / 3 = 1812.6666666667 (the remainder is 2, so 3 is not a divisor of 5438)
...
5438 / 5437 = 1.0001839249586 (the remainder is 1, so 5437 is not a divisor of 5438)
5438 / 5438 = 1 (the remainder is 0, so 5438 is a divisor of 5438)

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 5438 (i.e. 73.742796258346). 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 $\frac{N}{d} = D$ )

5438 / 1 = 5438 (the remainder is 0, so 1 and 5438 are divisors of 5438)
5438 / 2 = 2719 (the remainder is 0, so 2 and 2719 are divisors of 5438)
5438 / 3 = 1812.6666666667 (the remainder is 2, so 3 is not a divisor of 5438)
...
5438 / 72 = 75.527777777778 (the remainder is 38, so 72 is not a divisor of 5438)
5438 / 73 = 74.493150684932 (the remainder is 36, so 73 is not a divisor of 5438)