What are the divisors of 3306?

1, 2, 3, 6, 19, 29, 38, 57, 58, 87, 114, 174, 551, 1102, 1653, 3306

There is a total of 16 positive divisors.
The sum of these divisors is 7200.
The arithmetic mean is 450.

8 even divisors

2, 6, 38, 58, 114, 174, 1102, 3306

8 odd divisors

1, 3, 19, 29, 57, 87, 551, 1653

How to compute the divisors of 3306?

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 3306 by each of the numbers from 1 to 3306 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)

3306 / 1 = 3306 (the remainder is 0, so 1 is a divisor of 3306)
3306 / 2 = 1653 (the remainder is 0, so 2 is a divisor of 3306)
3306 / 3 = 1102 (the remainder is 0, so 3 is a divisor of 3306)
...
3306 / 3305 = 1.0003025718608 (the remainder is 1, so 3305 is not a divisor of 3306)
3306 / 3306 = 1 (the remainder is 0, so 3306 is a divisor of 3306)

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 3306 (i.e. 57.49782604586). 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 $\frac{N}{d} = D$ )

3306 / 1 = 3306 (the remainder is 0, so 1 and 3306 are divisors of 3306)
3306 / 2 = 1653 (the remainder is 0, so 2 and 1653 are divisors of 3306)
3306 / 3 = 1102 (the remainder is 0, so 3 and 1102 are divisors of 3306)
...
3306 / 56 = 59.035714285714 (the remainder is 2, so 56 is not a divisor of 3306)
3306 / 57 = 58 (the remainder is 0, so 57 and 58 are divisors of 3306)