What are the divisors of 1881?

1, 3, 9, 11, 19, 33, 57, 99, 171, 209, 627, 1881

There is a total of 12 positive divisors.
The sum of these divisors is 3120.
The arithmetic mean is 260.

12 odd divisors

1, 3, 9, 11, 19, 33, 57, 99, 171, 209, 627, 1881

How to compute the divisors of 1881?

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 1881 by each of the numbers from 1 to 1881 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)

1881 / 1 = 1881 (the remainder is 0, so 1 is a divisor of 1881)
1881 / 2 = 940.5 (the remainder is 1, so 2 is not a divisor of 1881)
1881 / 3 = 627 (the remainder is 0, so 3 is a divisor of 1881)
...
1881 / 1880 = 1.0005319148936 (the remainder is 1, so 1880 is not a divisor of 1881)
1881 / 1881 = 1 (the remainder is 0, so 1881 is a divisor of 1881)

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 1881 (i.e. 43.370496884403). 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 $\frac{N}{d} = D$ )

1881 / 1 = 1881 (the remainder is 0, so 1 and 1881 are divisors of 1881)
1881 / 2 = 940.5 (the remainder is 1, so 2 is not a divisor of 1881)
1881 / 3 = 627 (the remainder is 0, so 3 and 627 are divisors of 1881)
...
1881 / 42 = 44.785714285714 (the remainder is 33, so 42 is not a divisor of 1881)
1881 / 43 = 43.744186046512 (the remainder is 32, so 43 is not a divisor of 1881)