What are the divisors of 5329?

1, 73, 5329

There is a total of 3 positive divisors.
The sum of these divisors is 5403.
The arithmetic mean is 1801.

3 odd divisors

1, 73, 5329

How to compute the divisors of 5329?

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 5329 by each of the numbers from 1 to 5329 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)

5329 / 1 = 5329 (the remainder is 0, so 1 is a divisor of 5329)
5329 / 2 = 2664.5 (the remainder is 1, so 2 is not a divisor of 5329)
5329 / 3 = 1776.3333333333 (the remainder is 1, so 3 is not a divisor of 5329)
...
5329 / 5328 = 1.0001876876877 (the remainder is 1, so 5328 is not a divisor of 5329)
5329 / 5329 = 1 (the remainder is 0, so 5329 is a divisor of 5329)

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 5329 (i.e. 73). 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 $\frac{N}{d} = D$ )

5329 / 1 = 5329 (the remainder is 0, so 1 and 5329 are divisors of 5329)
5329 / 2 = 2664.5 (the remainder is 1, so 2 is not a divisor of 5329)
5329 / 3 = 1776.3333333333 (the remainder is 1, so 3 is not a divisor of 5329)
...
5329 / 72 = 74.013888888889 (the remainder is 1, so 72 is not a divisor of 5329)
5329 / 73 = 73 (the remainder is 0, so 73 and 73 are divisors of 5329)