What are the divisors of 3627?

1, 3, 9, 13, 31, 39, 93, 117, 279, 403, 1209, 3627

There is a total of 12 positive divisors.
The sum of these divisors is 5824.
The arithmetic mean is 485.33333333333.

12 odd divisors

1, 3, 9, 13, 31, 39, 93, 117, 279, 403, 1209, 3627

How to compute the divisors of 3627?

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 3627 by each of the numbers from 1 to 3627 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)

3627 / 1 = 3627 (the remainder is 0, so 1 is a divisor of 3627)
3627 / 2 = 1813.5 (the remainder is 1, so 2 is not a divisor of 3627)
3627 / 3 = 1209 (the remainder is 0, so 3 is a divisor of 3627)
...
3627 / 3626 = 1.0002757859901 (the remainder is 1, so 3626 is not a divisor of 3627)
3627 / 3627 = 1 (the remainder is 0, so 3627 is a divisor of 3627)

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 3627 (i.e. 60.224579699654). 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 $\frac{N}{d} = D$ )

3627 / 1 = 3627 (the remainder is 0, so 1 and 3627 are divisors of 3627)
3627 / 2 = 1813.5 (the remainder is 1, so 2 is not a divisor of 3627)
3627 / 3 = 1209 (the remainder is 0, so 3 and 1209 are divisors of 3627)
...
3627 / 59 = 61.474576271186 (the remainder is 28, so 59 is not a divisor of 3627)
3627 / 60 = 60.45 (the remainder is 27, so 60 is not a divisor of 3627)