What are the divisors of 321?

1, 3, 107, 321

There is a total of 4 positive divisors.
The sum of these divisors is 432.
The arithmetic mean is 108.

4 odd divisors

1, 3, 107, 321

How to compute the divisors of 321?

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 321 by each of the numbers from 1 to 321 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)

321 / 1 = 321 (the remainder is 0, so 1 is a divisor of 321)
321 / 2 = 160.5 (the remainder is 1, so 2 is not a divisor of 321)
321 / 3 = 107 (the remainder is 0, so 3 is a divisor of 321)
...
321 / 320 = 1.003125 (the remainder is 1, so 320 is not a divisor of 321)
321 / 321 = 1 (the remainder is 0, so 321 is a divisor of 321)

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 321 (i.e. 17.916472867169). 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 $\frac{N}{d} = D$ )

321 / 1 = 321 (the remainder is 0, so 1 and 321 are divisors of 321)
321 / 2 = 160.5 (the remainder is 1, so 2 is not a divisor of 321)
321 / 3 = 107 (the remainder is 0, so 3 and 107 are divisors of 321)
...
321 / 16 = 20.0625 (the remainder is 1, so 16 is not a divisor of 321)
321 / 17 = 18.882352941176 (the remainder is 15, so 17 is not a divisor of 321)