What are the divisors of 2003?

1, 2003

There is a total of 2 positive divisors.
The sum of these divisors is 2004.
The arithmetic mean is 1002.

2 odd divisors

1, 2003

How to compute the divisors of 2003?

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 2003 by each of the numbers from 1 to 2003 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)

2003 / 1 = 2003 (the remainder is 0, so 1 is a divisor of 2003)
2003 / 2 = 1001.5 (the remainder is 1, so 2 is not a divisor of 2003)
2003 / 3 = 667.66666666667 (the remainder is 2, so 3 is not a divisor of 2003)
...
2003 / 2002 = 1.0004995004995 (the remainder is 1, so 2002 is not a divisor of 2003)
2003 / 2003 = 1 (the remainder is 0, so 2003 is a divisor of 2003)

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 2003 (i.e. 44.7548880012). 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 $\frac{N}{d} = D$ )

2003 / 1 = 2003 (the remainder is 0, so 1 and 2003 are divisors of 2003)
2003 / 2 = 1001.5 (the remainder is 1, so 2 is not a divisor of 2003)
2003 / 3 = 667.66666666667 (the remainder is 2, so 3 is not a divisor of 2003)
...
2003 / 43 = 46.581395348837 (the remainder is 25, so 43 is not a divisor of 2003)
2003 / 44 = 45.522727272727 (the remainder is 23, so 44 is not a divisor of 2003)