What are the divisors of 1833?

1, 3, 13, 39, 47, 141, 611, 1833

There is a total of 8 positive divisors.
The sum of these divisors is 2688.
The arithmetic mean is 336.

8 odd divisors

1, 3, 13, 39, 47, 141, 611, 1833

How to compute the divisors of 1833?

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 1833 by each of the numbers from 1 to 1833 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)

1833 / 1 = 1833 (the remainder is 0, so 1 is a divisor of 1833)
1833 / 2 = 916.5 (the remainder is 1, so 2 is not a divisor of 1833)
1833 / 3 = 611 (the remainder is 0, so 3 is a divisor of 1833)
...
1833 / 1832 = 1.0005458515284 (the remainder is 1, so 1832 is not a divisor of 1833)
1833 / 1833 = 1 (the remainder is 0, so 1833 is a divisor of 1833)

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 1833 (i.e. 42.813549257215). 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 $\frac{N}{d} = D$ )

1833 / 1 = 1833 (the remainder is 0, so 1 and 1833 are divisors of 1833)
1833 / 2 = 916.5 (the remainder is 1, so 2 is not a divisor of 1833)
1833 / 3 = 611 (the remainder is 0, so 3 and 611 are divisors of 1833)
...
1833 / 41 = 44.707317073171 (the remainder is 29, so 41 is not a divisor of 1833)
1833 / 42 = 43.642857142857 (the remainder is 27, so 42 is not a divisor of 1833)