What are the divisors of 5910?

1, 2, 3, 5, 6, 10, 15, 30, 197, 394, 591, 985, 1182, 1970, 2955, 5910

There is a total of 16 positive divisors.
The sum of these divisors is 14256.
The arithmetic mean is 891.

8 even divisors

2, 6, 10, 30, 394, 1182, 1970, 5910

8 odd divisors

1, 3, 5, 15, 197, 591, 985, 2955

How to compute the divisors of 5910?

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 5910 by each of the numbers from 1 to 5910 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)

5910 / 1 = 5910 (the remainder is 0, so 1 is a divisor of 5910)
5910 / 2 = 2955 (the remainder is 0, so 2 is a divisor of 5910)
5910 / 3 = 1970 (the remainder is 0, so 3 is a divisor of 5910)
...
5910 / 5909 = 1.0001692333728 (the remainder is 1, so 5909 is not a divisor of 5910)
5910 / 5910 = 1 (the remainder is 0, so 5910 is a divisor of 5910)

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 5910 (i.e. 76.87652437513). 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 $\frac{N}{d} = D$ )

5910 / 1 = 5910 (the remainder is 0, so 1 and 5910 are divisors of 5910)
5910 / 2 = 2955 (the remainder is 0, so 2 and 2955 are divisors of 5910)
5910 / 3 = 1970 (the remainder is 0, so 3 and 1970 are divisors of 5910)
...
5910 / 75 = 78.8 (the remainder is 60, so 75 is not a divisor of 5910)
5910 / 76 = 77.763157894737 (the remainder is 58, so 76 is not a divisor of 5910)