What are the divisors of 1824?

1, 2, 3, 4, 6, 8, 12, 16, 19, 24, 32, 38, 48, 57, 76, 96, 114, 152, 228, 304, 456, 608, 912, 1824

There is a total of 24 positive divisors.
The sum of these divisors is 5040.
The arithmetic mean is 210.

20 even divisors

2, 4, 6, 8, 12, 16, 24, 32, 38, 48, 76, 96, 114, 152, 228, 304, 456, 608, 912, 1824

4 odd divisors

1, 3, 19, 57

How to compute the divisors of 1824?

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 1824 by each of the numbers from 1 to 1824 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)

1824 / 1 = 1824 (the remainder is 0, so 1 is a divisor of 1824)
1824 / 2 = 912 (the remainder is 0, so 2 is a divisor of 1824)
1824 / 3 = 608 (the remainder is 0, so 3 is a divisor of 1824)
...
1824 / 1823 = 1.0005485463522 (the remainder is 1, so 1823 is not a divisor of 1824)
1824 / 1824 = 1 (the remainder is 0, so 1824 is a divisor of 1824)

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 1824 (i.e. 42.708313008125). 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 $\frac{N}{d} = D$ )

1824 / 1 = 1824 (the remainder is 0, so 1 and 1824 are divisors of 1824)
1824 / 2 = 912 (the remainder is 0, so 2 and 912 are divisors of 1824)
1824 / 3 = 608 (the remainder is 0, so 3 and 608 are divisors of 1824)
...
1824 / 41 = 44.487804878049 (the remainder is 20, so 41 is not a divisor of 1824)
1824 / 42 = 43.428571428571 (the remainder is 18, so 42 is not a divisor of 1824)