What are the divisors of 5535?

1, 3, 5, 9, 15, 27, 41, 45, 123, 135, 205, 369, 615, 1107, 1845, 5535

There is a total of 16 positive divisors.
The sum of these divisors is 10080.
The arithmetic mean is 630.

16 odd divisors

1, 3, 5, 9, 15, 27, 41, 45, 123, 135, 205, 369, 615, 1107, 1845, 5535

How to compute the divisors of 5535?

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 5535 by each of the numbers from 1 to 5535 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)

5535 / 1 = 5535 (the remainder is 0, so 1 is a divisor of 5535)
5535 / 2 = 2767.5 (the remainder is 1, so 2 is not a divisor of 5535)
5535 / 3 = 1845 (the remainder is 0, so 3 is a divisor of 5535)
...
5535 / 5534 = 1.0001807011203 (the remainder is 1, so 5534 is not a divisor of 5535)
5535 / 5535 = 1 (the remainder is 0, so 5535 is a divisor of 5535)

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 5535 (i.e. 74.397580605823). 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 $\frac{N}{d} = D$ )

5535 / 1 = 5535 (the remainder is 0, so 1 and 5535 are divisors of 5535)
5535 / 2 = 2767.5 (the remainder is 1, so 2 is not a divisor of 5535)
5535 / 3 = 1845 (the remainder is 0, so 3 and 1845 are divisors of 5535)
...
5535 / 73 = 75.821917808219 (the remainder is 60, so 73 is not a divisor of 5535)
5535 / 74 = 74.797297297297 (the remainder is 59, so 74 is not a divisor of 5535)