What are the divisors of 5306?

1, 2, 7, 14, 379, 758, 2653, 5306

There is a total of 8 positive divisors.
The sum of these divisors is 9120.
The arithmetic mean is 1140.

4 even divisors

2, 14, 758, 5306

4 odd divisors

1, 7, 379, 2653

How to compute the divisors of 5306?

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 5306 by each of the numbers from 1 to 5306 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)

5306 / 1 = 5306 (the remainder is 0, so 1 is a divisor of 5306)
5306 / 2 = 2653 (the remainder is 0, so 2 is a divisor of 5306)
5306 / 3 = 1768.6666666667 (the remainder is 2, so 3 is not a divisor of 5306)
...
5306 / 5305 = 1.0001885014138 (the remainder is 1, so 5305 is not a divisor of 5306)
5306 / 5306 = 1 (the remainder is 0, so 5306 is a divisor of 5306)

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 5306 (i.e. 72.842295405897). 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 $\frac{N}{d} = D$ )

5306 / 1 = 5306 (the remainder is 0, so 1 and 5306 are divisors of 5306)
5306 / 2 = 2653 (the remainder is 0, so 2 and 2653 are divisors of 5306)
5306 / 3 = 1768.6666666667 (the remainder is 2, so 3 is not a divisor of 5306)
...
5306 / 71 = 74.732394366197 (the remainder is 52, so 71 is not a divisor of 5306)
5306 / 72 = 73.694444444444 (the remainder is 50, so 72 is not a divisor of 5306)