What are the divisors of 3612?

1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 43, 84, 86, 129, 172, 258, 301, 516, 602, 903, 1204, 1806, 3612

There is a total of 24 positive divisors.
The sum of these divisors is 9856.
The arithmetic mean is 410.66666666667.

16 even divisors

2, 4, 6, 12, 14, 28, 42, 84, 86, 172, 258, 516, 602, 1204, 1806, 3612

8 odd divisors

1, 3, 7, 21, 43, 129, 301, 903

How to compute the divisors of 3612?

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 3612 by each of the numbers from 1 to 3612 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)

3612 / 1 = 3612 (the remainder is 0, so 1 is a divisor of 3612)
3612 / 2 = 1806 (the remainder is 0, so 2 is a divisor of 3612)
3612 / 3 = 1204 (the remainder is 0, so 3 is a divisor of 3612)
...
3612 / 3611 = 1.0002769315979 (the remainder is 1, so 3611 is not a divisor of 3612)
3612 / 3612 = 1 (the remainder is 0, so 3612 is a divisor of 3612)

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 3612 (i.e. 60.099916805267). 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 $\frac{N}{d} = D$ )

3612 / 1 = 3612 (the remainder is 0, so 1 and 3612 are divisors of 3612)
3612 / 2 = 1806 (the remainder is 0, so 2 and 1806 are divisors of 3612)
3612 / 3 = 1204 (the remainder is 0, so 3 and 1204 are divisors of 3612)
...
3612 / 59 = 61.220338983051 (the remainder is 13, so 59 is not a divisor of 3612)
3612 / 60 = 60.2 (the remainder is 12, so 60 is not a divisor of 3612)