What are the divisors of 5993?

1, 13, 461, 5993

There is a total of 4 positive divisors.
The sum of these divisors is 6468.
The arithmetic mean is 1617.

4 odd divisors

1, 13, 461, 5993

How to compute the divisors of 5993?

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 5993 by each of the numbers from 1 to 5993 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)

5993 / 1 = 5993 (the remainder is 0, so 1 is a divisor of 5993)
5993 / 2 = 2996.5 (the remainder is 1, so 2 is not a divisor of 5993)
5993 / 3 = 1997.6666666667 (the remainder is 2, so 3 is not a divisor of 5993)
...
5993 / 5992 = 1.0001668891856 (the remainder is 1, so 5992 is not a divisor of 5993)
5993 / 5993 = 1 (the remainder is 0, so 5993 is a divisor of 5993)

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 5993 (i.e. 77.414468931848). 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 $\frac{N}{d} = D$ )

5993 / 1 = 5993 (the remainder is 0, so 1 and 5993 are divisors of 5993)
5993 / 2 = 2996.5 (the remainder is 1, so 2 is not a divisor of 5993)
5993 / 3 = 1997.6666666667 (the remainder is 2, so 3 is not a divisor of 5993)
...
5993 / 76 = 78.855263157895 (the remainder is 65, so 76 is not a divisor of 5993)
5993 / 77 = 77.831168831169 (the remainder is 64, so 77 is not a divisor of 5993)