What are the divisors of 9204?

1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 59, 78, 118, 156, 177, 236, 354, 708, 767, 1534, 2301, 3068, 4602, 9204

There is a total of 24 positive divisors.
The sum of these divisors is 23520.
The arithmetic mean is 980.

16 even divisors

2, 4, 6, 12, 26, 52, 78, 118, 156, 236, 354, 708, 1534, 3068, 4602, 9204

8 odd divisors

1, 3, 13, 39, 59, 177, 767, 2301

How to compute the divisors of 9204?

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 9204 by each of the numbers from 1 to 9204 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)

9204 / 1 = 9204 (the remainder is 0, so 1 is a divisor of 9204)
9204 / 2 = 4602 (the remainder is 0, so 2 is a divisor of 9204)
9204 / 3 = 3068 (the remainder is 0, so 3 is a divisor of 9204)
...
9204 / 9203 = 1.0001086602195 (the remainder is 1, so 9203 is not a divisor of 9204)
9204 / 9204 = 1 (the remainder is 0, so 9204 is a divisor of 9204)

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 9204 (i.e. 95.937479641692). 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 $\frac{N}{d} = D$ )

9204 / 1 = 9204 (the remainder is 0, so 1 and 9204 are divisors of 9204)
9204 / 2 = 4602 (the remainder is 0, so 2 and 4602 are divisors of 9204)
9204 / 3 = 3068 (the remainder is 0, so 3 and 3068 are divisors of 9204)
...
9204 / 94 = 97.914893617021 (the remainder is 86, so 94 is not a divisor of 9204)
9204 / 95 = 96.884210526316 (the remainder is 84, so 95 is not a divisor of 9204)