What are the divisors of 8080?

1, 2, 4, 5, 8, 10, 16, 20, 40, 80, 101, 202, 404, 505, 808, 1010, 1616, 2020, 4040, 8080

There is a total of 20 positive divisors.
The sum of these divisors is 18972.
The arithmetic mean is 948.6.

16 even divisors

2, 4, 8, 10, 16, 20, 40, 80, 202, 404, 808, 1010, 1616, 2020, 4040, 8080

4 odd divisors

1, 5, 101, 505

How to compute the divisors of 8080?

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 8080 by each of the numbers from 1 to 8080 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)

8080 / 1 = 8080 (the remainder is 0, so 1 is a divisor of 8080)
8080 / 2 = 4040 (the remainder is 0, so 2 is a divisor of 8080)
8080 / 3 = 2693.3333333333 (the remainder is 1, so 3 is not a divisor of 8080)
...
8080 / 8079 = 1.0001237776953 (the remainder is 1, so 8079 is not a divisor of 8080)
8080 / 8080 = 1 (the remainder is 0, so 8080 is a divisor of 8080)

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 8080 (i.e. 89.888820216977). 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 $\frac{N}{d} = D$ )

8080 / 1 = 8080 (the remainder is 0, so 1 and 8080 are divisors of 8080)
8080 / 2 = 4040 (the remainder is 0, so 2 and 4040 are divisors of 8080)
8080 / 3 = 2693.3333333333 (the remainder is 1, so 3 is not a divisor of 8080)
...
8080 / 88 = 91.818181818182 (the remainder is 72, so 88 is not a divisor of 8080)
8080 / 89 = 90.786516853933 (the remainder is 70, so 89 is not a divisor of 8080)