What are the divisors of 3384?

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 47, 72, 94, 141, 188, 282, 376, 423, 564, 846, 1128, 1692, 3384

There is a total of 24 positive divisors.
The sum of these divisors is 9360.
The arithmetic mean is 390.

18 even divisors

2, 4, 6, 8, 12, 18, 24, 36, 72, 94, 188, 282, 376, 564, 846, 1128, 1692, 3384

6 odd divisors

1, 3, 9, 47, 141, 423

How to compute the divisors of 3384?

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 3384 by each of the numbers from 1 to 3384 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)

3384 / 1 = 3384 (the remainder is 0, so 1 is a divisor of 3384)
3384 / 2 = 1692 (the remainder is 0, so 2 is a divisor of 3384)
3384 / 3 = 1128 (the remainder is 0, so 3 is a divisor of 3384)
...
3384 / 3383 = 1.0002955956252 (the remainder is 1, so 3383 is not a divisor of 3384)
3384 / 3384 = 1 (the remainder is 0, so 3384 is a divisor of 3384)

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 3384 (i.e. 58.172158288996). 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 $\frac{N}{d} = D$ )

3384 / 1 = 3384 (the remainder is 0, so 1 and 3384 are divisors of 3384)
3384 / 2 = 1692 (the remainder is 0, so 2 and 1692 are divisors of 3384)
3384 / 3 = 1128 (the remainder is 0, so 3 and 1128 are divisors of 3384)
...
3384 / 57 = 59.368421052632 (the remainder is 21, so 57 is not a divisor of 3384)
3384 / 58 = 58.344827586207 (the remainder is 20, so 58 is not a divisor of 3384)