What are the divisors of 3624?

1, 2, 3, 4, 6, 8, 12, 24, 151, 302, 453, 604, 906, 1208, 1812, 3624

There is a total of 16 positive divisors.
The sum of these divisors is 9120.
The arithmetic mean is 570.

12 even divisors

2, 4, 6, 8, 12, 24, 302, 604, 906, 1208, 1812, 3624

4 odd divisors

1, 3, 151, 453

How to compute the divisors of 3624?

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 3624 by each of the numbers from 1 to 3624 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)

3624 / 1 = 3624 (the remainder is 0, so 1 is a divisor of 3624)
3624 / 2 = 1812 (the remainder is 0, so 2 is a divisor of 3624)
3624 / 3 = 1208 (the remainder is 0, so 3 is a divisor of 3624)
...
3624 / 3623 = 1.0002760143527 (the remainder is 1, so 3623 is not a divisor of 3624)
3624 / 3624 = 1 (the remainder is 0, so 3624 is a divisor of 3624)

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 3624 (i.e. 60.19966777317). 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 $\frac{N}{d} = D$ )

3624 / 1 = 3624 (the remainder is 0, so 1 and 3624 are divisors of 3624)
3624 / 2 = 1812 (the remainder is 0, so 2 and 1812 are divisors of 3624)
3624 / 3 = 1208 (the remainder is 0, so 3 and 1208 are divisors of 3624)
...
3624 / 59 = 61.423728813559 (the remainder is 25, so 59 is not a divisor of 3624)
3624 / 60 = 60.4 (the remainder is 24, so 60 is not a divisor of 3624)