What are the divisors of 5490?

1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 61, 90, 122, 183, 305, 366, 549, 610, 915, 1098, 1830, 2745, 5490

There is a total of 24 positive divisors.
The sum of these divisors is 14508.
The arithmetic mean is 604.5.

12 even divisors

2, 6, 10, 18, 30, 90, 122, 366, 610, 1098, 1830, 5490

12 odd divisors

1, 3, 5, 9, 15, 45, 61, 183, 305, 549, 915, 2745

How to compute the divisors of 5490?

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 5490 by each of the numbers from 1 to 5490 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)

5490 / 1 = 5490 (the remainder is 0, so 1 is a divisor of 5490)
5490 / 2 = 2745 (the remainder is 0, so 2 is a divisor of 5490)
5490 / 3 = 1830 (the remainder is 0, so 3 is a divisor of 5490)
...
5490 / 5489 = 1.0001821825469 (the remainder is 1, so 5489 is not a divisor of 5490)
5490 / 5490 = 1 (the remainder is 0, so 5490 is a divisor of 5490)

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 5490 (i.e. 74.094534211371). 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 $\frac{N}{d} = D$ )

5490 / 1 = 5490 (the remainder is 0, so 1 and 5490 are divisors of 5490)
5490 / 2 = 2745 (the remainder is 0, so 2 and 2745 are divisors of 5490)
5490 / 3 = 1830 (the remainder is 0, so 3 and 1830 are divisors of 5490)
...
5490 / 73 = 75.205479452055 (the remainder is 15, so 73 is not a divisor of 5490)
5490 / 74 = 74.189189189189 (the remainder is 14, so 74 is not a divisor of 5490)