What are the divisors of 1989?

1, 3, 9, 13, 17, 39, 51, 117, 153, 221, 663, 1989

There is a total of 12 positive divisors.
The sum of these divisors is 3276.
The arithmetic mean is 273.

12 odd divisors

1, 3, 9, 13, 17, 39, 51, 117, 153, 221, 663, 1989

How to compute the divisors of 1989?

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 1989 by each of the numbers from 1 to 1989 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)

1989 / 1 = 1989 (the remainder is 0, so 1 is a divisor of 1989)
1989 / 2 = 994.5 (the remainder is 1, so 2 is not a divisor of 1989)
1989 / 3 = 663 (the remainder is 0, so 3 is a divisor of 1989)
...
1989 / 1988 = 1.0005030181087 (the remainder is 1, so 1988 is not a divisor of 1989)
1989 / 1989 = 1 (the remainder is 0, so 1989 is a divisor of 1989)

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 1989 (i.e. 44.598206241956). 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 $\frac{N}{d} = D$ )

1989 / 1 = 1989 (the remainder is 0, so 1 and 1989 are divisors of 1989)
1989 / 2 = 994.5 (the remainder is 1, so 2 is not a divisor of 1989)
1989 / 3 = 663 (the remainder is 0, so 3 and 663 are divisors of 1989)
...
1989 / 43 = 46.255813953488 (the remainder is 11, so 43 is not a divisor of 1989)
1989 / 44 = 45.204545454545 (the remainder is 9, so 44 is not a divisor of 1989)