What are the divisors of 2020?

1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020

There is a total of 12 positive divisors.
The sum of these divisors is 4284.
The arithmetic mean is 357.

8 even divisors

2, 4, 10, 20, 202, 404, 1010, 2020

4 odd divisors

1, 5, 101, 505

How to compute the divisors of 2020?

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 2020 by each of the numbers from 1 to 2020 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)

2020 / 1 = 2020 (the remainder is 0, so 1 is a divisor of 2020)
2020 / 2 = 1010 (the remainder is 0, so 2 is a divisor of 2020)
2020 / 3 = 673.33333333333 (the remainder is 1, so 3 is not a divisor of 2020)
...
2020 / 2019 = 1.0004952947003 (the remainder is 1, so 2019 is not a divisor of 2020)
2020 / 2020 = 1 (the remainder is 0, so 2020 is a divisor of 2020)

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 2020 (i.e. 44.944410108488). 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 $\frac{N}{d} = D$ )

2020 / 1 = 2020 (the remainder is 0, so 1 and 2020 are divisors of 2020)
2020 / 2 = 1010 (the remainder is 0, so 2 and 1010 are divisors of 2020)
2020 / 3 = 673.33333333333 (the remainder is 1, so 3 is not a divisor of 2020)
...
2020 / 43 = 46.976744186047 (the remainder is 42, so 43 is not a divisor of 2020)
2020 / 44 = 45.909090909091 (the remainder is 40, so 44 is not a divisor of 2020)