What are the divisors of 3440?

1, 2, 4, 5, 8, 10, 16, 20, 40, 43, 80, 86, 172, 215, 344, 430, 688, 860, 1720, 3440

There is a total of 20 positive divisors.
The sum of these divisors is 8184.
The arithmetic mean is 409.2.

16 even divisors

2, 4, 8, 10, 16, 20, 40, 80, 86, 172, 344, 430, 688, 860, 1720, 3440

4 odd divisors

1, 5, 43, 215

How to compute the divisors of 3440?

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 3440 by each of the numbers from 1 to 3440 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)

3440 / 1 = 3440 (the remainder is 0, so 1 is a divisor of 3440)
3440 / 2 = 1720 (the remainder is 0, so 2 is a divisor of 3440)
3440 / 3 = 1146.6666666667 (the remainder is 2, so 3 is not a divisor of 3440)
...
3440 / 3439 = 1.0002907822041 (the remainder is 1, so 3439 is not a divisor of 3440)
3440 / 3440 = 1 (the remainder is 0, so 3440 is a divisor of 3440)

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 3440 (i.e. 58.651513194461). 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 $\frac{N}{d} = D$ )

3440 / 1 = 3440 (the remainder is 0, so 1 and 3440 are divisors of 3440)
3440 / 2 = 1720 (the remainder is 0, so 2 and 1720 are divisors of 3440)
3440 / 3 = 1146.6666666667 (the remainder is 2, so 3 is not a divisor of 3440)
...
3440 / 57 = 60.350877192982 (the remainder is 20, so 57 is not a divisor of 3440)
3440 / 58 = 59.310344827586 (the remainder is 18, so 58 is not a divisor of 3440)