What are the divisors of 1839?

1, 3, 613, 1839

There is a total of 4 positive divisors.
The sum of these divisors is 2456.
The arithmetic mean is 614.

4 odd divisors

1, 3, 613, 1839

How to compute the divisors of 1839?

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 1839 by each of the numbers from 1 to 1839 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)

1839 / 1 = 1839 (the remainder is 0, so 1 is a divisor of 1839)
1839 / 2 = 919.5 (the remainder is 1, so 2 is not a divisor of 1839)
1839 / 3 = 613 (the remainder is 0, so 3 is a divisor of 1839)
...
1839 / 1838 = 1.0005440696409 (the remainder is 1, so 1838 is not a divisor of 1839)
1839 / 1839 = 1 (the remainder is 0, so 1839 is a divisor of 1839)

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 1839 (i.e. 42.883563284783). 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 $\frac{N}{d} = D$ )

1839 / 1 = 1839 (the remainder is 0, so 1 and 1839 are divisors of 1839)
1839 / 2 = 919.5 (the remainder is 1, so 2 is not a divisor of 1839)
1839 / 3 = 613 (the remainder is 0, so 3 and 613 are divisors of 1839)
...
1839 / 41 = 44.853658536585 (the remainder is 35, so 41 is not a divisor of 1839)
1839 / 42 = 43.785714285714 (the remainder is 33, so 42 is not a divisor of 1839)