What are the divisors of 4120?

1, 2, 4, 5, 8, 10, 20, 40, 103, 206, 412, 515, 824, 1030, 2060, 4120

There is a total of 16 positive divisors.
The sum of these divisors is 9360.
The arithmetic mean is 585.

12 even divisors

2, 4, 8, 10, 20, 40, 206, 412, 824, 1030, 2060, 4120

4 odd divisors

1, 5, 103, 515

How to compute the divisors of 4120?

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 4120 by each of the numbers from 1 to 4120 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)

4120 / 1 = 4120 (the remainder is 0, so 1 is a divisor of 4120)
4120 / 2 = 2060 (the remainder is 0, so 2 is a divisor of 4120)
4120 / 3 = 1373.3333333333 (the remainder is 1, so 3 is not a divisor of 4120)
...
4120 / 4119 = 1.0002427773731 (the remainder is 1, so 4119 is not a divisor of 4120)
4120 / 4120 = 1 (the remainder is 0, so 4120 is a divisor of 4120)

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 4120 (i.e. 64.187226143525). 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 $\frac{N}{d} = D$ )

4120 / 1 = 4120 (the remainder is 0, so 1 and 4120 are divisors of 4120)
4120 / 2 = 2060 (the remainder is 0, so 2 and 2060 are divisors of 4120)
4120 / 3 = 1373.3333333333 (the remainder is 1, so 3 is not a divisor of 4120)
...
4120 / 63 = 65.396825396825 (the remainder is 25, so 63 is not a divisor of 4120)
4120 / 64 = 64.375 (the remainder is 24, so 64 is not a divisor of 4120)