What are the divisors of 204?

1, 2, 3, 4, 6, 12, 17, 34, 51, 68, 102, 204

There is a total of 12 positive divisors.
The sum of these divisors is 504.
The arithmetic mean is 42.

8 even divisors

2, 4, 6, 12, 34, 68, 102, 204

4 odd divisors

1, 3, 17, 51

How to compute the divisors of 204?

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 204 by each of the numbers from 1 to 204 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)

204 / 1 = 204 (the remainder is 0, so 1 is a divisor of 204)
204 / 2 = 102 (the remainder is 0, so 2 is a divisor of 204)
204 / 3 = 68 (the remainder is 0, so 3 is a divisor of 204)
...
204 / 203 = 1.0049261083744 (the remainder is 1, so 203 is not a divisor of 204)
204 / 204 = 1 (the remainder is 0, so 204 is a divisor of 204)

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 204 (i.e. 14.282856857086). 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 $\frac{N}{d} = D$ )

204 / 1 = 204 (the remainder is 0, so 1 and 204 are divisors of 204)
204 / 2 = 102 (the remainder is 0, so 2 and 102 are divisors of 204)
204 / 3 = 68 (the remainder is 0, so 3 and 68 are divisors of 204)
...
204 / 13 = 15.692307692308 (the remainder is 9, so 13 is not a divisor of 204)
204 / 14 = 14.571428571429 (the remainder is 8, so 14 is not a divisor of 204)