What are the divisors of 2002?

1, 2, 7, 11, 13, 14, 22, 26, 77, 91, 143, 154, 182, 286, 1001, 2002

There is a total of 16 positive divisors.
The sum of these divisors is 4032.
The arithmetic mean is 252.

8 even divisors

2, 14, 22, 26, 154, 182, 286, 2002

8 odd divisors

1, 7, 11, 13, 77, 91, 143, 1001

How to compute the divisors of 2002?

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 2002 by each of the numbers from 1 to 2002 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)

2002 / 1 = 2002 (the remainder is 0, so 1 is a divisor of 2002)
2002 / 2 = 1001 (the remainder is 0, so 2 is a divisor of 2002)
2002 / 3 = 667.33333333333 (the remainder is 1, so 3 is not a divisor of 2002)
...
2002 / 2001 = 1.0004997501249 (the remainder is 1, so 2001 is not a divisor of 2002)
2002 / 2002 = 1 (the remainder is 0, so 2002 is a divisor of 2002)

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 2002 (i.e. 44.743714642394). 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 $\frac{N}{d} = D$ )

2002 / 1 = 2002 (the remainder is 0, so 1 and 2002 are divisors of 2002)
2002 / 2 = 1001 (the remainder is 0, so 2 and 1001 are divisors of 2002)
2002 / 3 = 667.33333333333 (the remainder is 1, so 3 is not a divisor of 2002)
...
2002 / 43 = 46.558139534884 (the remainder is 24, so 43 is not a divisor of 2002)
2002 / 44 = 45.5 (the remainder is 22, so 44 is not a divisor of 2002)