What are the divisors of 3912?

1, 2, 3, 4, 6, 8, 12, 24, 163, 326, 489, 652, 978, 1304, 1956, 3912

There is a total of 16 positive divisors.
The sum of these divisors is 9840.
The arithmetic mean is 615.

12 even divisors

2, 4, 6, 8, 12, 24, 326, 652, 978, 1304, 1956, 3912

4 odd divisors

1, 3, 163, 489

How to compute the divisors of 3912?

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 3912 by each of the numbers from 1 to 3912 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)

3912 / 1 = 3912 (the remainder is 0, so 1 is a divisor of 3912)
3912 / 2 = 1956 (the remainder is 0, so 2 is a divisor of 3912)
3912 / 3 = 1304 (the remainder is 0, so 3 is a divisor of 3912)
...
3912 / 3911 = 1.0002556890821 (the remainder is 1, so 3911 is not a divisor of 3912)
3912 / 3912 = 1 (the remainder is 0, so 3912 is a divisor of 3912)

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 3912 (i.e. 62.545983084448). 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 $\frac{N}{d} = D$ )

3912 / 1 = 3912 (the remainder is 0, so 1 and 3912 are divisors of 3912)
3912 / 2 = 1956 (the remainder is 0, so 2 and 1956 are divisors of 3912)
3912 / 3 = 1304 (the remainder is 0, so 3 and 1304 are divisors of 3912)
...
3912 / 61 = 64.131147540984 (the remainder is 8, so 61 is not a divisor of 3912)
3912 / 62 = 63.096774193548 (the remainder is 6, so 62 is not a divisor of 3912)