What are the divisors of 1803?

1, 3, 601, 1803

There is a total of 4 positive divisors.
The sum of these divisors is 2408.
The arithmetic mean is 602.

4 odd divisors

1, 3, 601, 1803

How to compute the divisors of 1803?

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 1803 by each of the numbers from 1 to 1803 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)

1803 / 1 = 1803 (the remainder is 0, so 1 is a divisor of 1803)
1803 / 2 = 901.5 (the remainder is 1, so 2 is not a divisor of 1803)
1803 / 3 = 601 (the remainder is 0, so 3 is a divisor of 1803)
...
1803 / 1802 = 1.0005549389567 (the remainder is 1, so 1802 is not a divisor of 1803)
1803 / 1803 = 1 (the remainder is 0, so 1803 is a divisor of 1803)

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 1803 (i.e. 42.461747491124). 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 $\frac{N}{d} = D$ )

1803 / 1 = 1803 (the remainder is 0, so 1 and 1803 are divisors of 1803)
1803 / 2 = 901.5 (the remainder is 1, so 2 is not a divisor of 1803)
1803 / 3 = 601 (the remainder is 0, so 3 and 601 are divisors of 1803)
...
1803 / 41 = 43.975609756098 (the remainder is 40, so 41 is not a divisor of 1803)
1803 / 42 = 42.928571428571 (the remainder is 39, so 42 is not a divisor of 1803)