What are the divisors of 5301?

1, 3, 9, 19, 31, 57, 93, 171, 279, 589, 1767, 5301

There is a total of 12 positive divisors.
The sum of these divisors is 8320.
The arithmetic mean is 693.33333333333.

12 odd divisors

1, 3, 9, 19, 31, 57, 93, 171, 279, 589, 1767, 5301

How to compute the divisors of 5301?

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 5301 by each of the numbers from 1 to 5301 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)

5301 / 1 = 5301 (the remainder is 0, so 1 is a divisor of 5301)
5301 / 2 = 2650.5 (the remainder is 1, so 2 is not a divisor of 5301)
5301 / 3 = 1767 (the remainder is 0, so 3 is a divisor of 5301)
...
5301 / 5300 = 1.0001886792453 (the remainder is 1, so 5300 is not a divisor of 5301)
5301 / 5301 = 1 (the remainder is 0, so 5301 is a divisor of 5301)

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 5301 (i.e. 72.80796659707). 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 $\frac{N}{d} = D$ )

5301 / 1 = 5301 (the remainder is 0, so 1 and 5301 are divisors of 5301)
5301 / 2 = 2650.5 (the remainder is 1, so 2 is not a divisor of 5301)
5301 / 3 = 1767 (the remainder is 0, so 3 and 1767 are divisors of 5301)
...
5301 / 71 = 74.661971830986 (the remainder is 47, so 71 is not a divisor of 5301)
5301 / 72 = 73.625 (the remainder is 45, so 72 is not a divisor of 5301)