What are the divisors of 4599?

1, 3, 7, 9, 21, 63, 73, 219, 511, 657, 1533, 4599

There is a total of 12 positive divisors.
The sum of these divisors is 7696.
The arithmetic mean is 641.33333333333.

12 odd divisors

1, 3, 7, 9, 21, 63, 73, 219, 511, 657, 1533, 4599

How to compute the divisors of 4599?

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 4599 by each of the numbers from 1 to 4599 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)

4599 / 1 = 4599 (the remainder is 0, so 1 is a divisor of 4599)
4599 / 2 = 2299.5 (the remainder is 1, so 2 is not a divisor of 4599)
4599 / 3 = 1533 (the remainder is 0, so 3 is a divisor of 4599)
...
4599 / 4598 = 1.0002174858634 (the remainder is 1, so 4598 is not a divisor of 4599)
4599 / 4599 = 1 (the remainder is 0, so 4599 is a divisor of 4599)

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 4599 (i.e. 67.815927332744). 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 $\frac{N}{d} = D$ )

4599 / 1 = 4599 (the remainder is 0, so 1 and 4599 are divisors of 4599)
4599 / 2 = 2299.5 (the remainder is 1, so 2 is not a divisor of 4599)
4599 / 3 = 1533 (the remainder is 0, so 3 and 1533 are divisors of 4599)
...
4599 / 66 = 69.681818181818 (the remainder is 45, so 66 is not a divisor of 4599)
4599 / 67 = 68.641791044776 (the remainder is 43, so 67 is not a divisor of 4599)