What are the divisors of 2001?

1, 3, 23, 29, 69, 87, 667, 2001

There is a total of 8 positive divisors.
The sum of these divisors is 2880.
The arithmetic mean is 360.

8 odd divisors

1, 3, 23, 29, 69, 87, 667, 2001

How to compute the divisors of 2001?

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 2001 by each of the numbers from 1 to 2001 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)

2001 / 1 = 2001 (the remainder is 0, so 1 is a divisor of 2001)
2001 / 2 = 1000.5 (the remainder is 1, so 2 is not a divisor of 2001)
2001 / 3 = 667 (the remainder is 0, so 3 is a divisor of 2001)
...
2001 / 2000 = 1.0005 (the remainder is 1, so 2000 is not a divisor of 2001)
2001 / 2001 = 1 (the remainder is 0, so 2001 is a divisor of 2001)

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 2001 (i.e. 44.73253849269). 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 $\frac{N}{d} = D$ )

2001 / 1 = 2001 (the remainder is 0, so 1 and 2001 are divisors of 2001)
2001 / 2 = 1000.5 (the remainder is 1, so 2 is not a divisor of 2001)
2001 / 3 = 667 (the remainder is 0, so 3 and 667 are divisors of 2001)
...
2001 / 43 = 46.53488372093 (the remainder is 23, so 43 is not a divisor of 2001)
2001 / 44 = 45.477272727273 (the remainder is 21, so 44 is not a divisor of 2001)