What are the divisors of 2108?

1, 2, 4, 17, 31, 34, 62, 68, 124, 527, 1054, 2108

There is a total of 12 positive divisors.
The sum of these divisors is 4032.
The arithmetic mean is 336.

8 even divisors

2, 4, 34, 62, 68, 124, 1054, 2108

4 odd divisors

1, 17, 31, 527

How to compute the divisors of 2108?

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 2108 by each of the numbers from 1 to 2108 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)

2108 / 1 = 2108 (the remainder is 0, so 1 is a divisor of 2108)
2108 / 2 = 1054 (the remainder is 0, so 2 is a divisor of 2108)
2108 / 3 = 702.66666666667 (the remainder is 2, so 3 is not a divisor of 2108)
...
2108 / 2107 = 1.000474608448 (the remainder is 1, so 2107 is not a divisor of 2108)
2108 / 2108 = 1 (the remainder is 0, so 2108 is a divisor of 2108)

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 2108 (i.e. 45.912961132996). 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 $\frac{N}{d} = D$ )

2108 / 1 = 2108 (the remainder is 0, so 1 and 2108 are divisors of 2108)
2108 / 2 = 1054 (the remainder is 0, so 2 and 1054 are divisors of 2108)
2108 / 3 = 702.66666666667 (the remainder is 2, so 3 is not a divisor of 2108)
...
2108 / 44 = 47.909090909091 (the remainder is 40, so 44 is not a divisor of 2108)
2108 / 45 = 46.844444444444 (the remainder is 38, so 45 is not a divisor of 2108)