What are the divisors of 3339?

1, 3, 7, 9, 21, 53, 63, 159, 371, 477, 1113, 3339

There is a total of 12 positive divisors.
The sum of these divisors is 5616.
The arithmetic mean is 468.

12 odd divisors

1, 3, 7, 9, 21, 53, 63, 159, 371, 477, 1113, 3339

How to compute the divisors of 3339?

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 3339 by each of the numbers from 1 to 3339 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)

3339 / 1 = 3339 (the remainder is 0, so 1 is a divisor of 3339)
3339 / 2 = 1669.5 (the remainder is 1, so 2 is not a divisor of 3339)
3339 / 3 = 1113 (the remainder is 0, so 3 is a divisor of 3339)
...
3339 / 3338 = 1.0002995805872 (the remainder is 1, so 3338 is not a divisor of 3339)
3339 / 3339 = 1 (the remainder is 0, so 3339 is a divisor of 3339)

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 3339 (i.e. 57.784080852775). 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 $\frac{N}{d} = D$ )

3339 / 1 = 3339 (the remainder is 0, so 1 and 3339 are divisors of 3339)
3339 / 2 = 1669.5 (the remainder is 1, so 2 is not a divisor of 3339)
3339 / 3 = 1113 (the remainder is 0, so 3 and 1113 are divisors of 3339)
...
3339 / 56 = 59.625 (the remainder is 35, so 56 is not a divisor of 3339)
3339 / 57 = 58.578947368421 (the remainder is 33, so 57 is not a divisor of 3339)