What are the divisors of 8050?

1, 2, 5, 7, 10, 14, 23, 25, 35, 46, 50, 70, 115, 161, 175, 230, 322, 350, 575, 805, 1150, 1610, 4025, 8050

There is a total of 24 positive divisors.
The sum of these divisors is 17856.
The arithmetic mean is 744.

12 even divisors

2, 10, 14, 46, 50, 70, 230, 322, 350, 1150, 1610, 8050

12 odd divisors

1, 5, 7, 23, 25, 35, 115, 161, 175, 575, 805, 4025

How to compute the divisors of 8050?

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 8050 by each of the numbers from 1 to 8050 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)

8050 / 1 = 8050 (the remainder is 0, so 1 is a divisor of 8050)
8050 / 2 = 4025 (the remainder is 0, so 2 is a divisor of 8050)
8050 / 3 = 2683.3333333333 (the remainder is 1, so 3 is not a divisor of 8050)
...
8050 / 8049 = 1.0001242390359 (the remainder is 1, so 8049 is not a divisor of 8050)
8050 / 8050 = 1 (the remainder is 0, so 8050 is a divisor of 8050)

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 8050 (i.e. 89.721792224632). 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 $\frac{N}{d} = D$ )

8050 / 1 = 8050 (the remainder is 0, so 1 and 8050 are divisors of 8050)
8050 / 2 = 4025 (the remainder is 0, so 2 and 4025 are divisors of 8050)
8050 / 3 = 2683.3333333333 (the remainder is 1, so 3 is not a divisor of 8050)
...
8050 / 88 = 91.477272727273 (the remainder is 42, so 88 is not a divisor of 8050)
8050 / 89 = 90.449438202247 (the remainder is 40, so 89 is not a divisor of 8050)