What are the divisors of 5906?

1, 2, 2953, 5906

There is a total of 4 positive divisors.
The sum of these divisors is 8862.
The arithmetic mean is 2215.5.

2 even divisors

2, 5906

2 odd divisors

1, 2953

How to compute the divisors of 5906?

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 5906 by each of the numbers from 1 to 5906 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)

5906 / 1 = 5906 (the remainder is 0, so 1 is a divisor of 5906)
5906 / 2 = 2953 (the remainder is 0, so 2 is a divisor of 5906)
5906 / 3 = 1968.6666666667 (the remainder is 2, so 3 is not a divisor of 5906)
...
5906 / 5905 = 1.0001693480102 (the remainder is 1, so 5905 is not a divisor of 5906)
5906 / 5906 = 1 (the remainder is 0, so 5906 is a divisor of 5906)

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 5906 (i.e. 76.850504227363). 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 $\frac{N}{d} = D$ )

5906 / 1 = 5906 (the remainder is 0, so 1 and 5906 are divisors of 5906)
5906 / 2 = 2953 (the remainder is 0, so 2 and 2953 are divisors of 5906)
5906 / 3 = 1968.6666666667 (the remainder is 2, so 3 is not a divisor of 5906)
...
5906 / 75 = 78.746666666667 (the remainder is 56, so 75 is not a divisor of 5906)
5906 / 76 = 77.710526315789 (the remainder is 54, so 76 is not a divisor of 5906)