What are the divisors of 399?

1, 3, 7, 19, 21, 57, 133, 399

There is a total of 8 positive divisors.
The sum of these divisors is 640.
The arithmetic mean is 80.

8 odd divisors

1, 3, 7, 19, 21, 57, 133, 399

How to compute the divisors of 399?

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 399 by each of the numbers from 1 to 399 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)

399 / 1 = 399 (the remainder is 0, so 1 is a divisor of 399)
399 / 2 = 199.5 (the remainder is 1, so 2 is not a divisor of 399)
399 / 3 = 133 (the remainder is 0, so 3 is a divisor of 399)
...
399 / 398 = 1.0025125628141 (the remainder is 1, so 398 is not a divisor of 399)
399 / 399 = 1 (the remainder is 0, so 399 is a divisor of 399)

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 399 (i.e. 19.974984355438). 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 $\frac{N}{d} = D$ )

399 / 1 = 399 (the remainder is 0, so 1 and 399 are divisors of 399)
399 / 2 = 199.5 (the remainder is 1, so 2 is not a divisor of 399)
399 / 3 = 133 (the remainder is 0, so 3 and 133 are divisors of 399)
...
399 / 18 = 22.166666666667 (the remainder is 3, so 18 is not a divisor of 399)
399 / 19 = 21 (the remainder is 0, so 19 and 21 are divisors of 399)