What are the divisors of 5336?

1, 2, 4, 8, 23, 29, 46, 58, 92, 116, 184, 232, 667, 1334, 2668, 5336

There is a total of 16 positive divisors.
The sum of these divisors is 10800.
The arithmetic mean is 675.

12 even divisors

2, 4, 8, 46, 58, 92, 116, 184, 232, 1334, 2668, 5336

4 odd divisors

1, 23, 29, 667

How to compute the divisors of 5336?

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 5336 by each of the numbers from 1 to 5336 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)

5336 / 1 = 5336 (the remainder is 0, so 1 is a divisor of 5336)
5336 / 2 = 2668 (the remainder is 0, so 2 is a divisor of 5336)
5336 / 3 = 1778.6666666667 (the remainder is 2, so 3 is not a divisor of 5336)
...
5336 / 5335 = 1.0001874414246 (the remainder is 1, so 5335 is not a divisor of 5336)
5336 / 5336 = 1 (the remainder is 0, so 5336 is a divisor of 5336)

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 5336 (i.e. 73.047929470999). 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 $\frac{N}{d} = D$ )

5336 / 1 = 5336 (the remainder is 0, so 1 and 5336 are divisors of 5336)
5336 / 2 = 2668 (the remainder is 0, so 2 and 2668 are divisors of 5336)
5336 / 3 = 1778.6666666667 (the remainder is 2, so 3 is not a divisor of 5336)
...
5336 / 72 = 74.111111111111 (the remainder is 8, so 72 is not a divisor of 5336)
5336 / 73 = 73.095890410959 (the remainder is 7, so 73 is not a divisor of 5336)