What are the divisors of 5840?

1, 2, 4, 5, 8, 10, 16, 20, 40, 73, 80, 146, 292, 365, 584, 730, 1168, 1460, 2920, 5840

There is a total of 20 positive divisors.
The sum of these divisors is 13764.
The arithmetic mean is 688.2.

16 even divisors

2, 4, 8, 10, 16, 20, 40, 80, 146, 292, 584, 730, 1168, 1460, 2920, 5840

4 odd divisors

1, 5, 73, 365

How to compute the divisors of 5840?

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 5840 by each of the numbers from 1 to 5840 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)

5840 / 1 = 5840 (the remainder is 0, so 1 is a divisor of 5840)
5840 / 2 = 2920 (the remainder is 0, so 2 is a divisor of 5840)
5840 / 3 = 1946.6666666667 (the remainder is 2, so 3 is not a divisor of 5840)
...
5840 / 5839 = 1.0001712622024 (the remainder is 1, so 5839 is not a divisor of 5840)
5840 / 5840 = 1 (the remainder is 0, so 5840 is a divisor of 5840)

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 5840 (i.e. 76.419892698171). 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 $\frac{N}{d} = D$ )

5840 / 1 = 5840 (the remainder is 0, so 1 and 5840 are divisors of 5840)
5840 / 2 = 2920 (the remainder is 0, so 2 and 2920 are divisors of 5840)
5840 / 3 = 1946.6666666667 (the remainder is 2, so 3 is not a divisor of 5840)
...
5840 / 75 = 77.866666666667 (the remainder is 65, so 75 is not a divisor of 5840)
5840 / 76 = 76.842105263158 (the remainder is 64, so 76 is not a divisor of 5840)