What are the divisors of 4002?

1, 2, 3, 6, 23, 29, 46, 58, 69, 87, 138, 174, 667, 1334, 2001, 4002

There is a total of 16 positive divisors.
The sum of these divisors is 8640.
The arithmetic mean is 540.

8 even divisors

2, 6, 46, 58, 138, 174, 1334, 4002

8 odd divisors

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

How to compute the divisors of 4002?

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 4002 by each of the numbers from 1 to 4002 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)

4002 / 1 = 4002 (the remainder is 0, so 1 is a divisor of 4002)
4002 / 2 = 2001 (the remainder is 0, so 2 is a divisor of 4002)
4002 / 3 = 1334 (the remainder is 0, so 3 is a divisor of 4002)
...
4002 / 4001 = 1.0002499375156 (the remainder is 1, so 4001 is not a divisor of 4002)
4002 / 4002 = 1 (the remainder is 0, so 4002 is a divisor of 4002)

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 4002 (i.e. 63.261362615739). 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 $\frac{N}{d} = D$ )

4002 / 1 = 4002 (the remainder is 0, so 1 and 4002 are divisors of 4002)
4002 / 2 = 2001 (the remainder is 0, so 2 and 2001 are divisors of 4002)
4002 / 3 = 1334 (the remainder is 0, so 3 and 1334 are divisors of 4002)
...
4002 / 62 = 64.548387096774 (the remainder is 34, so 62 is not a divisor of 4002)
4002 / 63 = 63.52380952381 (the remainder is 33, so 63 is not a divisor of 4002)