What are the divisors of 5907?

1, 3, 11, 33, 179, 537, 1969, 5907

There is a total of 8 positive divisors.
The sum of these divisors is 8640.
The arithmetic mean is 1080.

8 odd divisors

1, 3, 11, 33, 179, 537, 1969, 5907

How to compute the divisors of 5907?

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 5907 by each of the numbers from 1 to 5907 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)

5907 / 1 = 5907 (the remainder is 0, so 1 is a divisor of 5907)
5907 / 2 = 2953.5 (the remainder is 1, so 2 is not a divisor of 5907)
5907 / 3 = 1969 (the remainder is 0, so 3 is a divisor of 5907)
...
5907 / 5906 = 1.0001693193363 (the remainder is 1, so 5906 is not a divisor of 5907)
5907 / 5907 = 1 (the remainder is 0, so 5907 is a divisor of 5907)

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 5907 (i.e. 76.857010090167). 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 $\frac{N}{d} = D$ )

5907 / 1 = 5907 (the remainder is 0, so 1 and 5907 are divisors of 5907)
5907 / 2 = 2953.5 (the remainder is 1, so 2 is not a divisor of 5907)
5907 / 3 = 1969 (the remainder is 0, so 3 and 1969 are divisors of 5907)
...
5907 / 75 = 78.76 (the remainder is 57, so 75 is not a divisor of 5907)
5907 / 76 = 77.723684210526 (the remainder is 55, so 76 is not a divisor of 5907)