What are the divisors of 1995?

1, 3, 5, 7, 15, 19, 21, 35, 57, 95, 105, 133, 285, 399, 665, 1995

There is a total of 16 positive divisors.
The sum of these divisors is 3840.
The arithmetic mean is 240.

16 odd divisors

1, 3, 5, 7, 15, 19, 21, 35, 57, 95, 105, 133, 285, 399, 665, 1995

How to compute the divisors of 1995?

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 1995 by each of the numbers from 1 to 1995 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)

1995 / 1 = 1995 (the remainder is 0, so 1 is a divisor of 1995)
1995 / 2 = 997.5 (the remainder is 1, so 2 is not a divisor of 1995)
1995 / 3 = 665 (the remainder is 0, so 3 is a divisor of 1995)
...
1995 / 1994 = 1.0005015045135 (the remainder is 1, so 1994 is not a divisor of 1995)
1995 / 1995 = 1 (the remainder is 0, so 1995 is a divisor of 1995)

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 1995 (i.e. 44.665422868255). 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 $\frac{N}{d} = D$ )

1995 / 1 = 1995 (the remainder is 0, so 1 and 1995 are divisors of 1995)
1995 / 2 = 997.5 (the remainder is 1, so 2 is not a divisor of 1995)
1995 / 3 = 665 (the remainder is 0, so 3 and 665 are divisors of 1995)
...
1995 / 43 = 46.395348837209 (the remainder is 17, so 43 is not a divisor of 1995)
1995 / 44 = 45.340909090909 (the remainder is 15, so 44 is not a divisor of 1995)