What are the divisors of 2840?

1, 2, 4, 5, 8, 10, 20, 40, 71, 142, 284, 355, 568, 710, 1420, 2840

There is a total of 16 positive divisors.
The sum of these divisors is 6480.
The arithmetic mean is 405.

12 even divisors

2, 4, 8, 10, 20, 40, 142, 284, 568, 710, 1420, 2840

4 odd divisors

1, 5, 71, 355

How to compute the divisors of 2840?

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 2840 by each of the numbers from 1 to 2840 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)

2840 / 1 = 2840 (the remainder is 0, so 1 is a divisor of 2840)
2840 / 2 = 1420 (the remainder is 0, so 2 is a divisor of 2840)
2840 / 3 = 946.66666666667 (the remainder is 2, so 3 is not a divisor of 2840)
...
2840 / 2839 = 1.0003522367031 (the remainder is 1, so 2839 is not a divisor of 2840)
2840 / 2840 = 1 (the remainder is 0, so 2840 is a divisor of 2840)

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 2840 (i.e. 53.291650377897). 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 $\frac{N}{d} = D$ )

2840 / 1 = 2840 (the remainder is 0, so 1 and 2840 are divisors of 2840)
2840 / 2 = 1420 (the remainder is 0, so 2 and 1420 are divisors of 2840)
2840 / 3 = 946.66666666667 (the remainder is 2, so 3 is not a divisor of 2840)
...
2840 / 52 = 54.615384615385 (the remainder is 32, so 52 is not a divisor of 2840)
2840 / 53 = 53.584905660377 (the remainder is 31, so 53 is not a divisor of 2840)