What are the divisors of 3412?

1, 2, 4, 853, 1706, 3412

There is a total of 6 positive divisors.
The sum of these divisors is 5978.
The arithmetic mean is 996.33333333333.

4 even divisors

2, 4, 1706, 3412

2 odd divisors

1, 853

How to compute the divisors of 3412?

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 3412 by each of the numbers from 1 to 3412 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)

3412 / 1 = 3412 (the remainder is 0, so 1 is a divisor of 3412)
3412 / 2 = 1706 (the remainder is 0, so 2 is a divisor of 3412)
3412 / 3 = 1137.3333333333 (the remainder is 1, so 3 is not a divisor of 3412)
...
3412 / 3411 = 1.0002931691586 (the remainder is 1, so 3411 is not a divisor of 3412)
3412 / 3412 = 1 (the remainder is 0, so 3412 is a divisor of 3412)

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 3412 (i.e. 58.412327466041). 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 $\frac{N}{d} = D$ )

3412 / 1 = 3412 (the remainder is 0, so 1 and 3412 are divisors of 3412)
3412 / 2 = 1706 (the remainder is 0, so 2 and 1706 are divisors of 3412)
3412 / 3 = 1137.3333333333 (the remainder is 1, so 3 is not a divisor of 3412)
...
3412 / 57 = 59.859649122807 (the remainder is 49, so 57 is not a divisor of 3412)
3412 / 58 = 58.827586206897 (the remainder is 48, so 58 is not a divisor of 3412)