What are the divisors of 2680?

1, 2, 4, 5, 8, 10, 20, 40, 67, 134, 268, 335, 536, 670, 1340, 2680

There is a total of 16 positive divisors.
The sum of these divisors is 6120.
The arithmetic mean is 382.5.

12 even divisors

2, 4, 8, 10, 20, 40, 134, 268, 536, 670, 1340, 2680

4 odd divisors

1, 5, 67, 335

How to compute the divisors of 2680?

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 2680 by each of the numbers from 1 to 2680 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)

2680 / 1 = 2680 (the remainder is 0, so 1 is a divisor of 2680)
2680 / 2 = 1340 (the remainder is 0, so 2 is a divisor of 2680)
2680 / 3 = 893.33333333333 (the remainder is 1, so 3 is not a divisor of 2680)
...
2680 / 2679 = 1.0003732736096 (the remainder is 1, so 2679 is not a divisor of 2680)
2680 / 2680 = 1 (the remainder is 0, so 2680 is a divisor of 2680)

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 2680 (i.e. 51.768716422179). 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 $\frac{N}{d} = D$ )

2680 / 1 = 2680 (the remainder is 0, so 1 and 2680 are divisors of 2680)
2680 / 2 = 1340 (the remainder is 0, so 2 and 1340 are divisors of 2680)
2680 / 3 = 893.33333333333 (the remainder is 1, so 3 is not a divisor of 2680)
...
2680 / 50 = 53.6 (the remainder is 30, so 50 is not a divisor of 2680)
2680 / 51 = 52.549019607843 (the remainder is 28, so 51 is not a divisor of 2680)