What are the divisors of 9234?

1, 2, 3, 6, 9, 18, 19, 27, 38, 54, 57, 81, 114, 162, 171, 243, 342, 486, 513, 1026, 1539, 3078, 4617, 9234

There is a total of 24 positive divisors.
The sum of these divisors is 21840.
The arithmetic mean is 910.

12 even divisors

2, 6, 18, 38, 54, 114, 162, 342, 486, 1026, 3078, 9234

12 odd divisors

1, 3, 9, 19, 27, 57, 81, 171, 243, 513, 1539, 4617

How to compute the divisors of 9234?

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 9234 by each of the numbers from 1 to 9234 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)

9234 / 1 = 9234 (the remainder is 0, so 1 is a divisor of 9234)
9234 / 2 = 4617 (the remainder is 0, so 2 is a divisor of 9234)
9234 / 3 = 3078 (the remainder is 0, so 3 is a divisor of 9234)
...
9234 / 9233 = 1.0001083071591 (the remainder is 1, so 9233 is not a divisor of 9234)
9234 / 9234 = 1 (the remainder is 0, so 9234 is a divisor of 9234)

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 9234 (i.e. 96.093704268282). 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 $\frac{N}{d} = D$ )

9234 / 1 = 9234 (the remainder is 0, so 1 and 9234 are divisors of 9234)
9234 / 2 = 4617 (the remainder is 0, so 2 and 4617 are divisors of 9234)
9234 / 3 = 3078 (the remainder is 0, so 3 and 3078 are divisors of 9234)
...
9234 / 95 = 97.2 (the remainder is 19, so 95 is not a divisor of 9234)
9234 / 96 = 96.1875 (the remainder is 18, so 96 is not a divisor of 9234)