What are the divisors of 3936?

1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 41, 48, 82, 96, 123, 164, 246, 328, 492, 656, 984, 1312, 1968, 3936

There is a total of 24 positive divisors.
The sum of these divisors is 10584.
The arithmetic mean is 441.

20 even divisors

2, 4, 6, 8, 12, 16, 24, 32, 48, 82, 96, 164, 246, 328, 492, 656, 984, 1312, 1968, 3936

4 odd divisors

1, 3, 41, 123

How to compute the divisors of 3936?

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 3936 by each of the numbers from 1 to 3936 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)

3936 / 1 = 3936 (the remainder is 0, so 1 is a divisor of 3936)
3936 / 2 = 1968 (the remainder is 0, so 2 is a divisor of 3936)
3936 / 3 = 1312 (the remainder is 0, so 3 is a divisor of 3936)
...
3936 / 3935 = 1.0002541296061 (the remainder is 1, so 3935 is not a divisor of 3936)
3936 / 3936 = 1 (the remainder is 0, so 3936 is a divisor of 3936)

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 3936 (i.e. 62.737548565432). 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 $\frac{N}{d} = D$ )

3936 / 1 = 3936 (the remainder is 0, so 1 and 3936 are divisors of 3936)
3936 / 2 = 1968 (the remainder is 0, so 2 and 1968 are divisors of 3936)
3936 / 3 = 1312 (the remainder is 0, so 3 and 1312 are divisors of 3936)
...
3936 / 61 = 64.524590163934 (the remainder is 32, so 61 is not a divisor of 3936)
3936 / 62 = 63.483870967742 (the remainder is 30, so 62 is not a divisor of 3936)