What are the divisors of 3220?

1, 2, 4, 5, 7, 10, 14, 20, 23, 28, 35, 46, 70, 92, 115, 140, 161, 230, 322, 460, 644, 805, 1610, 3220

There is a total of 24 positive divisors.
The sum of these divisors is 8064.
The arithmetic mean is 336.

16 even divisors

2, 4, 10, 14, 20, 28, 46, 70, 92, 140, 230, 322, 460, 644, 1610, 3220

8 odd divisors

1, 5, 7, 23, 35, 115, 161, 805

How to compute the divisors of 3220?

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 3220 by each of the numbers from 1 to 3220 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)

3220 / 1 = 3220 (the remainder is 0, so 1 is a divisor of 3220)
3220 / 2 = 1610 (the remainder is 0, so 2 is a divisor of 3220)
3220 / 3 = 1073.3333333333 (the remainder is 1, so 3 is not a divisor of 3220)
...
3220 / 3219 = 1.0003106554831 (the remainder is 1, so 3219 is not a divisor of 3220)
3220 / 3220 = 1 (the remainder is 0, so 3220 is a divisor of 3220)

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 3220 (i.e. 56.745043836444). 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 $\frac{N}{d} = D$ )

3220 / 1 = 3220 (the remainder is 0, so 1 and 3220 are divisors of 3220)
3220 / 2 = 1610 (the remainder is 0, so 2 and 1610 are divisors of 3220)
3220 / 3 = 1073.3333333333 (the remainder is 1, so 3 is not a divisor of 3220)
...
3220 / 55 = 58.545454545455 (the remainder is 30, so 55 is not a divisor of 3220)
3220 / 56 = 57.5 (the remainder is 28, so 56 is not a divisor of 3220)