What are the divisors of 5308?

1, 2, 4, 1327, 2654, 5308

There is a total of 6 positive divisors.
The sum of these divisors is 9296.
The arithmetic mean is 1549.3333333333.

4 even divisors

2, 4, 2654, 5308

2 odd divisors

1, 1327

How to compute the divisors of 5308?

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 5308 by each of the numbers from 1 to 5308 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)

5308 / 1 = 5308 (the remainder is 0, so 1 is a divisor of 5308)
5308 / 2 = 2654 (the remainder is 0, so 2 is a divisor of 5308)
5308 / 3 = 1769.3333333333 (the remainder is 1, so 3 is not a divisor of 5308)
...
5308 / 5307 = 1.000188430375 (the remainder is 1, so 5307 is not a divisor of 5308)
5308 / 5308 = 1 (the remainder is 0, so 5308 is a divisor of 5308)

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 5308 (i.e. 72.856022400348). 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 $\frac{N}{d} = D$ )

5308 / 1 = 5308 (the remainder is 0, so 1 and 5308 are divisors of 5308)
5308 / 2 = 2654 (the remainder is 0, so 2 and 2654 are divisors of 5308)
5308 / 3 = 1769.3333333333 (the remainder is 1, so 3 is not a divisor of 5308)
...
5308 / 71 = 74.760563380282 (the remainder is 54, so 71 is not a divisor of 5308)
5308 / 72 = 73.722222222222 (the remainder is 52, so 72 is not a divisor of 5308)