What are the divisors of 4329?

1, 3, 9, 13, 37, 39, 111, 117, 333, 481, 1443, 4329

There is a total of 12 positive divisors.
The sum of these divisors is 6916.
The arithmetic mean is 576.33333333333.

12 odd divisors

1, 3, 9, 13, 37, 39, 111, 117, 333, 481, 1443, 4329

How to compute the divisors of 4329?

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 4329 by each of the numbers from 1 to 4329 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)

4329 / 1 = 4329 (the remainder is 0, so 1 is a divisor of 4329)
4329 / 2 = 2164.5 (the remainder is 1, so 2 is not a divisor of 4329)
4329 / 3 = 1443 (the remainder is 0, so 3 is a divisor of 4329)
...
4329 / 4328 = 1.0002310536044 (the remainder is 1, so 4328 is not a divisor of 4329)
4329 / 4329 = 1 (the remainder is 0, so 4329 is a divisor of 4329)

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 4329 (i.e. 65.795136598384). 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 $\frac{N}{d} = D$ )

4329 / 1 = 4329 (the remainder is 0, so 1 and 4329 are divisors of 4329)
4329 / 2 = 2164.5 (the remainder is 1, so 2 is not a divisor of 4329)
4329 / 3 = 1443 (the remainder is 0, so 3 and 1443 are divisors of 4329)
...
4329 / 64 = 67.640625 (the remainder is 41, so 64 is not a divisor of 4329)
4329 / 65 = 66.6 (the remainder is 39, so 65 is not a divisor of 4329)