What are the divisors of 3924?

1, 2, 3, 4, 6, 9, 12, 18, 36, 109, 218, 327, 436, 654, 981, 1308, 1962, 3924

There is a total of 18 positive divisors.
The sum of these divisors is 10010.
The arithmetic mean is 556.11111111111.

12 even divisors

2, 4, 6, 12, 18, 36, 218, 436, 654, 1308, 1962, 3924

6 odd divisors

1, 3, 9, 109, 327, 981

How to compute the divisors of 3924?

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 3924 by each of the numbers from 1 to 3924 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)

3924 / 1 = 3924 (the remainder is 0, so 1 is a divisor of 3924)
3924 / 2 = 1962 (the remainder is 0, so 2 is a divisor of 3924)
3924 / 3 = 1308 (the remainder is 0, so 3 is a divisor of 3924)
...
3924 / 3923 = 1.000254906959 (the remainder is 1, so 3923 is not a divisor of 3924)
3924 / 3924 = 1 (the remainder is 0, so 3924 is a divisor of 3924)

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 3924 (i.e. 62.641839053463). 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 $\frac{N}{d} = D$ )

3924 / 1 = 3924 (the remainder is 0, so 1 and 3924 are divisors of 3924)
3924 / 2 = 1962 (the remainder is 0, so 2 and 1962 are divisors of 3924)
3924 / 3 = 1308 (the remainder is 0, so 3 and 1308 are divisors of 3924)
...
3924 / 61 = 64.327868852459 (the remainder is 20, so 61 is not a divisor of 3924)
3924 / 62 = 63.290322580645 (the remainder is 18, so 62 is not a divisor of 3924)