What are the divisors of 5499?

1, 3, 9, 13, 39, 47, 117, 141, 423, 611, 1833, 5499

There is a total of 12 positive divisors.
The sum of these divisors is 8736.
The arithmetic mean is 728.

12 odd divisors

1, 3, 9, 13, 39, 47, 117, 141, 423, 611, 1833, 5499

How to compute the divisors of 5499?

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 5499 by each of the numbers from 1 to 5499 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)

5499 / 1 = 5499 (the remainder is 0, so 1 is a divisor of 5499)
5499 / 2 = 2749.5 (the remainder is 1, so 2 is not a divisor of 5499)
5499 / 3 = 1833 (the remainder is 0, so 3 is a divisor of 5499)
...
5499 / 5498 = 1.0001818843216 (the remainder is 1, so 5498 is not a divisor of 5499)
5499 / 5499 = 1 (the remainder is 0, so 5499 is a divisor of 5499)

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 5499 (i.e. 74.15524256585). 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 $\frac{N}{d} = D$ )

5499 / 1 = 5499 (the remainder is 0, so 1 and 5499 are divisors of 5499)
5499 / 2 = 2749.5 (the remainder is 1, so 2 is not a divisor of 5499)
5499 / 3 = 1833 (the remainder is 0, so 3 and 1833 are divisors of 5499)
...
5499 / 73 = 75.328767123288 (the remainder is 24, so 73 is not a divisor of 5499)
5499 / 74 = 74.310810810811 (the remainder is 23, so 74 is not a divisor of 5499)