What are the divisors of 5931?

1, 3, 9, 659, 1977, 5931

There is a total of 6 positive divisors.
The sum of these divisors is 8580.
The arithmetic mean is 1430.

6 odd divisors

1, 3, 9, 659, 1977, 5931

How to compute the divisors of 5931?

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 5931 by each of the numbers from 1 to 5931 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)

5931 / 1 = 5931 (the remainder is 0, so 1 is a divisor of 5931)
5931 / 2 = 2965.5 (the remainder is 1, so 2 is not a divisor of 5931)
5931 / 3 = 1977 (the remainder is 0, so 3 is a divisor of 5931)
...
5931 / 5930 = 1.0001686340641 (the remainder is 1, so 5930 is not a divisor of 5931)
5931 / 5931 = 1 (the remainder is 0, so 5931 is a divisor of 5931)

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 5931 (i.e. 77.012985917961). 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 $\frac{N}{d} = D$ )

5931 / 1 = 5931 (the remainder is 0, so 1 and 5931 are divisors of 5931)
5931 / 2 = 2965.5 (the remainder is 1, so 2 is not a divisor of 5931)
5931 / 3 = 1977 (the remainder is 0, so 3 and 1977 are divisors of 5931)
...
5931 / 76 = 78.039473684211 (the remainder is 3, so 76 is not a divisor of 5931)
5931 / 77 = 77.025974025974 (the remainder is 2, so 77 is not a divisor of 5931)