What are the divisors of 3404?

1, 2, 4, 23, 37, 46, 74, 92, 148, 851, 1702, 3404

There is a total of 12 positive divisors.
The sum of these divisors is 6384.
The arithmetic mean is 532.

8 even divisors

2, 4, 46, 74, 92, 148, 1702, 3404

4 odd divisors

1, 23, 37, 851

How to compute the divisors of 3404?

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 3404 by each of the numbers from 1 to 3404 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)

3404 / 1 = 3404 (the remainder is 0, so 1 is a divisor of 3404)
3404 / 2 = 1702 (the remainder is 0, so 2 is a divisor of 3404)
3404 / 3 = 1134.6666666667 (the remainder is 2, so 3 is not a divisor of 3404)
...
3404 / 3403 = 1.0002938583603 (the remainder is 1, so 3403 is not a divisor of 3404)
3404 / 3404 = 1 (the remainder is 0, so 3404 is a divisor of 3404)

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 3404 (i.e. 58.343808583259). 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 $\frac{N}{d} = D$ )

3404 / 1 = 3404 (the remainder is 0, so 1 and 3404 are divisors of 3404)
3404 / 2 = 1702 (the remainder is 0, so 2 and 1702 are divisors of 3404)
3404 / 3 = 1134.6666666667 (the remainder is 2, so 3 is not a divisor of 3404)
...
3404 / 57 = 59.719298245614 (the remainder is 41, so 57 is not a divisor of 3404)
3404 / 58 = 58.689655172414 (the remainder is 40, so 58 is not a divisor of 3404)