What are the divisors of 3630?

1, 2, 3, 5, 6, 10, 11, 15, 22, 30, 33, 55, 66, 110, 121, 165, 242, 330, 363, 605, 726, 1210, 1815, 3630

There is a total of 24 positive divisors.
The sum of these divisors is 9576.
The arithmetic mean is 399.

12 even divisors

2, 6, 10, 22, 30, 66, 110, 242, 330, 726, 1210, 3630

12 odd divisors

1, 3, 5, 11, 15, 33, 55, 121, 165, 363, 605, 1815

How to compute the divisors of 3630?

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 3630 by each of the numbers from 1 to 3630 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)

3630 / 1 = 3630 (the remainder is 0, so 1 is a divisor of 3630)
3630 / 2 = 1815 (the remainder is 0, so 2 is a divisor of 3630)
3630 / 3 = 1210 (the remainder is 0, so 3 is a divisor of 3630)
...
3630 / 3629 = 1.000275558005 (the remainder is 1, so 3629 is not a divisor of 3630)
3630 / 3630 = 1 (the remainder is 0, so 3630 is a divisor of 3630)

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 3630 (i.e. 60.249481325568). 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 $\frac{N}{d} = D$ )

3630 / 1 = 3630 (the remainder is 0, so 1 and 3630 are divisors of 3630)
3630 / 2 = 1815 (the remainder is 0, so 2 and 1815 are divisors of 3630)
3630 / 3 = 1210 (the remainder is 0, so 3 and 1210 are divisors of 3630)
...
3630 / 59 = 61.525423728814 (the remainder is 31, so 59 is not a divisor of 3630)
3630 / 60 = 60.5 (the remainder is 30, so 60 is not a divisor of 3630)