What are the divisors of 2007?

1, 3, 9, 223, 669, 2007

There is a total of 6 positive divisors.
The sum of these divisors is 2912.
The arithmetic mean is 485.33333333333.

6 odd divisors

1, 3, 9, 223, 669, 2007

How to compute the divisors of 2007?

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 2007 by each of the numbers from 1 to 2007 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)

2007 / 1 = 2007 (the remainder is 0, so 1 is a divisor of 2007)
2007 / 2 = 1003.5 (the remainder is 1, so 2 is not a divisor of 2007)
2007 / 3 = 669 (the remainder is 0, so 3 is a divisor of 2007)
...
2007 / 2006 = 1.0004985044865 (the remainder is 1, so 2006 is not a divisor of 2007)
2007 / 2007 = 1 (the remainder is 0, so 2007 is a divisor of 2007)

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 2007 (i.e. 44.799553569204). 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 $\frac{N}{d} = D$ )

2007 / 1 = 2007 (the remainder is 0, so 1 and 2007 are divisors of 2007)
2007 / 2 = 1003.5 (the remainder is 1, so 2 is not a divisor of 2007)
2007 / 3 = 669 (the remainder is 0, so 3 and 669 are divisors of 2007)
...
2007 / 43 = 46.674418604651 (the remainder is 29, so 43 is not a divisor of 2007)
2007 / 44 = 45.613636363636 (the remainder is 27, so 44 is not a divisor of 2007)