What are the divisors of 4040?

1, 2, 4, 5, 8, 10, 20, 40, 101, 202, 404, 505, 808, 1010, 2020, 4040

There is a total of 16 positive divisors.
The sum of these divisors is 9180.
The arithmetic mean is 573.75.

12 even divisors

2, 4, 8, 10, 20, 40, 202, 404, 808, 1010, 2020, 4040

4 odd divisors

1, 5, 101, 505

How to compute the divisors of 4040?

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 4040 by each of the numbers from 1 to 4040 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)

4040 / 1 = 4040 (the remainder is 0, so 1 is a divisor of 4040)
4040 / 2 = 2020 (the remainder is 0, so 2 is a divisor of 4040)
4040 / 3 = 1346.6666666667 (the remainder is 2, so 3 is not a divisor of 4040)
...
4040 / 4039 = 1.0002475860361 (the remainder is 1, so 4039 is not a divisor of 4040)
4040 / 4040 = 1 (the remainder is 0, so 4040 is a divisor of 4040)

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 4040 (i.e. 63.560994328283). 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 $\frac{N}{d} = D$ )

4040 / 1 = 4040 (the remainder is 0, so 1 and 4040 are divisors of 4040)
4040 / 2 = 2020 (the remainder is 0, so 2 and 2020 are divisors of 4040)
4040 / 3 = 1346.6666666667 (the remainder is 2, so 3 is not a divisor of 4040)
...
4040 / 62 = 65.161290322581 (the remainder is 10, so 62 is not a divisor of 4040)
4040 / 63 = 64.126984126984 (the remainder is 8, so 63 is not a divisor of 4040)