What are the divisors of 3267?

1, 3, 9, 11, 27, 33, 99, 121, 297, 363, 1089, 3267

There is a total of 12 positive divisors.
The sum of these divisors is 5320.
The arithmetic mean is 443.33333333333.

12 odd divisors

1, 3, 9, 11, 27, 33, 99, 121, 297, 363, 1089, 3267

How to compute the divisors of 3267?

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 3267 by each of the numbers from 1 to 3267 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)

3267 / 1 = 3267 (the remainder is 0, so 1 is a divisor of 3267)
3267 / 2 = 1633.5 (the remainder is 1, so 2 is not a divisor of 3267)
3267 / 3 = 1089 (the remainder is 0, so 3 is a divisor of 3267)
...
3267 / 3266 = 1.0003061849357 (the remainder is 1, so 3266 is not a divisor of 3267)
3267 / 3267 = 1 (the remainder is 0, so 3267 is a divisor of 3267)

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 3267 (i.e. 57.157676649773). 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 $\frac{N}{d} = D$ )

3267 / 1 = 3267 (the remainder is 0, so 1 and 3267 are divisors of 3267)
3267 / 2 = 1633.5 (the remainder is 1, so 2 is not a divisor of 3267)
3267 / 3 = 1089 (the remainder is 0, so 3 and 1089 are divisors of 3267)
...
3267 / 56 = 58.339285714286 (the remainder is 19, so 56 is not a divisor of 3267)
3267 / 57 = 57.315789473684 (the remainder is 18, so 57 is not a divisor of 3267)