What are the divisors of 3408?

1, 2, 3, 4, 6, 8, 12, 16, 24, 48, 71, 142, 213, 284, 426, 568, 852, 1136, 1704, 3408

There is a total of 20 positive divisors.
The sum of these divisors is 8928.
The arithmetic mean is 446.4.

16 even divisors

2, 4, 6, 8, 12, 16, 24, 48, 142, 284, 426, 568, 852, 1136, 1704, 3408

4 odd divisors

1, 3, 71, 213

How to compute the divisors of 3408?

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 3408 by each of the numbers from 1 to 3408 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)

3408 / 1 = 3408 (the remainder is 0, so 1 is a divisor of 3408)
3408 / 2 = 1704 (the remainder is 0, so 2 is a divisor of 3408)
3408 / 3 = 1136 (the remainder is 0, so 3 is a divisor of 3408)
...
3408 / 3407 = 1.0002935133549 (the remainder is 1, so 3407 is not a divisor of 3408)
3408 / 3408 = 1 (the remainder is 0, so 3408 is a divisor of 3408)

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 3408 (i.e. 58.378078077306). 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 $\frac{N}{d} = D$ )

3408 / 1 = 3408 (the remainder is 0, so 1 and 3408 are divisors of 3408)
3408 / 2 = 1704 (the remainder is 0, so 2 and 1704 are divisors of 3408)
3408 / 3 = 1136 (the remainder is 0, so 3 and 1136 are divisors of 3408)
...
3408 / 57 = 59.789473684211 (the remainder is 45, so 57 is not a divisor of 3408)
3408 / 58 = 58.758620689655 (the remainder is 44, so 58 is not a divisor of 3408)