What are the divisors of 3366?

1, 2, 3, 6, 9, 11, 17, 18, 22, 33, 34, 51, 66, 99, 102, 153, 187, 198, 306, 374, 561, 1122, 1683, 3366

There is a total of 24 positive divisors.
The sum of these divisors is 8424.
The arithmetic mean is 351.

12 even divisors

2, 6, 18, 22, 34, 66, 102, 198, 306, 374, 1122, 3366

12 odd divisors

1, 3, 9, 11, 17, 33, 51, 99, 153, 187, 561, 1683

How to compute the divisors of 3366?

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 3366 by each of the numbers from 1 to 3366 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)

3366 / 1 = 3366 (the remainder is 0, so 1 is a divisor of 3366)
3366 / 2 = 1683 (the remainder is 0, so 2 is a divisor of 3366)
3366 / 3 = 1122 (the remainder is 0, so 3 is a divisor of 3366)
...
3366 / 3365 = 1.0002971768202 (the remainder is 1, so 3365 is not a divisor of 3366)
3366 / 3366 = 1 (the remainder is 0, so 3366 is a divisor of 3366)

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 3366 (i.e. 58.017238817441). 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 $\frac{N}{d} = D$ )

3366 / 1 = 3366 (the remainder is 0, so 1 and 3366 are divisors of 3366)
3366 / 2 = 1683 (the remainder is 0, so 2 and 1683 are divisors of 3366)
3366 / 3 = 1122 (the remainder is 0, so 3 and 1122 are divisors of 3366)
...
3366 / 57 = 59.052631578947 (the remainder is 3, so 57 is not a divisor of 3366)
3366 / 58 = 58.034482758621 (the remainder is 2, so 58 is not a divisor of 3366)