What are the divisors of 1968?

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

There is a total of 20 positive divisors.
The sum of these divisors is 5208.
The arithmetic mean is 260.4.

16 even divisors

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

4 odd divisors

1, 3, 41, 123

How to compute the divisors of 1968?

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 1968 by each of the numbers from 1 to 1968 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)

1968 / 1 = 1968 (the remainder is 0, so 1 is a divisor of 1968)
1968 / 2 = 984 (the remainder is 0, so 2 is a divisor of 1968)
1968 / 3 = 656 (the remainder is 0, so 3 is a divisor of 1968)
...
1968 / 1967 = 1.0005083884087 (the remainder is 1, so 1967 is not a divisor of 1968)
1968 / 1968 = 1 (the remainder is 0, so 1968 is a divisor of 1968)

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 1968 (i.e. 44.362146025638). 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 $\frac{N}{d} = D$ )

1968 / 1 = 1968 (the remainder is 0, so 1 and 1968 are divisors of 1968)
1968 / 2 = 984 (the remainder is 0, so 2 and 984 are divisors of 1968)
1968 / 3 = 656 (the remainder is 0, so 3 and 656 are divisors of 1968)
...
1968 / 43 = 45.767441860465 (the remainder is 33, so 43 is not a divisor of 1968)
1968 / 44 = 44.727272727273 (the remainder is 32, so 44 is not a divisor of 1968)