What are the divisors of 2850?

1, 2, 3, 5, 6, 10, 15, 19, 25, 30, 38, 50, 57, 75, 95, 114, 150, 190, 285, 475, 570, 950, 1425, 2850

There is a total of 24 positive divisors.
The sum of these divisors is 7440.
The arithmetic mean is 310.

12 even divisors

2, 6, 10, 30, 38, 50, 114, 150, 190, 570, 950, 2850

12 odd divisors

1, 3, 5, 15, 19, 25, 57, 75, 95, 285, 475, 1425

How to compute the divisors of 2850?

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 2850 by each of the numbers from 1 to 2850 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)

2850 / 1 = 2850 (the remainder is 0, so 1 is a divisor of 2850)
2850 / 2 = 1425 (the remainder is 0, so 2 is a divisor of 2850)
2850 / 3 = 950 (the remainder is 0, so 3 is a divisor of 2850)
...
2850 / 2849 = 1.000351000351 (the remainder is 1, so 2849 is not a divisor of 2850)
2850 / 2850 = 1 (the remainder is 0, so 2850 is a divisor of 2850)

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 2850 (i.e. 53.385391260157). 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 $\frac{N}{d} = D$ )

2850 / 1 = 2850 (the remainder is 0, so 1 and 2850 are divisors of 2850)
2850 / 2 = 1425 (the remainder is 0, so 2 and 1425 are divisors of 2850)
2850 / 3 = 950 (the remainder is 0, so 3 and 950 are divisors of 2850)
...
2850 / 52 = 54.807692307692 (the remainder is 42, so 52 is not a divisor of 2850)
2850 / 53 = 53.77358490566 (the remainder is 41, so 53 is not a divisor of 2850)