What are the divisors of 318?

1, 2, 3, 6, 53, 106, 159, 318

There is a total of 8 positive divisors.
The sum of these divisors is 648.
The arithmetic mean is 81.

4 even divisors

2, 6, 106, 318

4 odd divisors

1, 3, 53, 159

How to compute the divisors of 318?

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 318 by each of the numbers from 1 to 318 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)

318 / 1 = 318 (the remainder is 0, so 1 is a divisor of 318)
318 / 2 = 159 (the remainder is 0, so 2 is a divisor of 318)
318 / 3 = 106 (the remainder is 0, so 3 is a divisor of 318)
...
318 / 317 = 1.0031545741325 (the remainder is 1, so 317 is not a divisor of 318)
318 / 318 = 1 (the remainder is 0, so 318 is a divisor of 318)

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 318 (i.e. 17.832554500127). 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 $\frac{N}{d} = D$ )

318 / 1 = 318 (the remainder is 0, so 1 and 318 are divisors of 318)
318 / 2 = 159 (the remainder is 0, so 2 and 159 are divisors of 318)
318 / 3 = 106 (the remainder is 0, so 3 and 106 are divisors of 318)
...
318 / 16 = 19.875 (the remainder is 14, so 16 is not a divisor of 318)
318 / 17 = 18.705882352941 (the remainder is 12, so 17 is not a divisor of 318)