What are the divisors of 1935?

1, 3, 5, 9, 15, 43, 45, 129, 215, 387, 645, 1935

There is a total of 12 positive divisors.
The sum of these divisors is 3432.
The arithmetic mean is 286.

12 odd divisors

1, 3, 5, 9, 15, 43, 45, 129, 215, 387, 645, 1935

How to compute the divisors of 1935?

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 1935 by each of the numbers from 1 to 1935 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)

1935 / 1 = 1935 (the remainder is 0, so 1 is a divisor of 1935)
1935 / 2 = 967.5 (the remainder is 1, so 2 is not a divisor of 1935)
1935 / 3 = 645 (the remainder is 0, so 3 is a divisor of 1935)
...
1935 / 1934 = 1.0005170630817 (the remainder is 1, so 1934 is not a divisor of 1935)
1935 / 1935 = 1 (the remainder is 0, so 1935 is a divisor of 1935)

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 1935 (i.e. 43.988634895846). 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 $\frac{N}{d} = D$ )

1935 / 1 = 1935 (the remainder is 0, so 1 and 1935 are divisors of 1935)
1935 / 2 = 967.5 (the remainder is 1, so 2 is not a divisor of 1935)
1935 / 3 = 645 (the remainder is 0, so 3 and 645 are divisors of 1935)
...
1935 / 42 = 46.071428571429 (the remainder is 3, so 42 is not a divisor of 1935)
1935 / 43 = 45 (the remainder is 0, so 43 and 45 are divisors of 1935)