What are the divisors of 5307?

1, 3, 29, 61, 87, 183, 1769, 5307

There is a total of 8 positive divisors.
The sum of these divisors is 7440.
The arithmetic mean is 930.

8 odd divisors

1, 3, 29, 61, 87, 183, 1769, 5307

How to compute the divisors of 5307?

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 5307 by each of the numbers from 1 to 5307 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)

5307 / 1 = 5307 (the remainder is 0, so 1 is a divisor of 5307)
5307 / 2 = 2653.5 (the remainder is 1, so 2 is not a divisor of 5307)
5307 / 3 = 1769 (the remainder is 0, so 3 is a divisor of 5307)
...
5307 / 5306 = 1.0001884658877 (the remainder is 1, so 5306 is not a divisor of 5307)
5307 / 5307 = 1 (the remainder is 0, so 5307 is a divisor of 5307)

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 5307 (i.e. 72.849159226445). 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 $\frac{N}{d} = D$ )

5307 / 1 = 5307 (the remainder is 0, so 1 and 5307 are divisors of 5307)
5307 / 2 = 2653.5 (the remainder is 1, so 2 is not a divisor of 5307)
5307 / 3 = 1769 (the remainder is 0, so 3 and 1769 are divisors of 5307)
...
5307 / 71 = 74.746478873239 (the remainder is 53, so 71 is not a divisor of 5307)
5307 / 72 = 73.708333333333 (the remainder is 51, so 72 is not a divisor of 5307)