What are the divisors of 9612?

1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 89, 108, 178, 267, 356, 534, 801, 1068, 1602, 2403, 3204, 4806, 9612

There is a total of 24 positive divisors.
The sum of these divisors is 25200.
The arithmetic mean is 1050.

16 even divisors

2, 4, 6, 12, 18, 36, 54, 108, 178, 356, 534, 1068, 1602, 3204, 4806, 9612

8 odd divisors

1, 3, 9, 27, 89, 267, 801, 2403

How to compute the divisors of 9612?

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 9612 by each of the numbers from 1 to 9612 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)

9612 / 1 = 9612 (the remainder is 0, so 1 is a divisor of 9612)
9612 / 2 = 4806 (the remainder is 0, so 2 is a divisor of 9612)
9612 / 3 = 3204 (the remainder is 0, so 3 is a divisor of 9612)
...
9612 / 9611 = 1.0001040474456 (the remainder is 1, so 9611 is not a divisor of 9612)
9612 / 9612 = 1 (the remainder is 0, so 9612 is a divisor of 9612)

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 9612 (i.e. 98.040807830209). 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 $\frac{N}{d} = D$ )

9612 / 1 = 9612 (the remainder is 0, so 1 and 9612 are divisors of 9612)
9612 / 2 = 4806 (the remainder is 0, so 2 and 4806 are divisors of 9612)
9612 / 3 = 3204 (the remainder is 0, so 3 and 3204 are divisors of 9612)
...
9612 / 97 = 99.092783505155 (the remainder is 9, so 97 is not a divisor of 9612)
9612 / 98 = 98.081632653061 (the remainder is 8, so 98 is not a divisor of 9612)