What are the divisors of 4005?

1, 3, 5, 9, 15, 45, 89, 267, 445, 801, 1335, 4005

There is a total of 12 positive divisors.
The sum of these divisors is 7020.
The arithmetic mean is 585.

12 odd divisors

1, 3, 5, 9, 15, 45, 89, 267, 445, 801, 1335, 4005

How to compute the divisors of 4005?

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 4005 by each of the numbers from 1 to 4005 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)

4005 / 1 = 4005 (the remainder is 0, so 1 is a divisor of 4005)
4005 / 2 = 2002.5 (the remainder is 1, so 2 is not a divisor of 4005)
4005 / 3 = 1335 (the remainder is 0, so 3 is a divisor of 4005)
...
4005 / 4004 = 1.0002497502498 (the remainder is 1, so 4004 is not a divisor of 4005)
4005 / 4005 = 1 (the remainder is 0, so 4005 is a divisor of 4005)

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 4005 (i.e. 63.285069329187). 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 $\frac{N}{d} = D$ )

4005 / 1 = 4005 (the remainder is 0, so 1 and 4005 are divisors of 4005)
4005 / 2 = 2002.5 (the remainder is 1, so 2 is not a divisor of 4005)
4005 / 3 = 1335 (the remainder is 0, so 3 and 1335 are divisors of 4005)
...
4005 / 62 = 64.596774193548 (the remainder is 37, so 62 is not a divisor of 4005)
4005 / 63 = 63.571428571429 (the remainder is 36, so 63 is not a divisor of 4005)