What are the divisors of 4095?

1, 3, 5, 7, 9, 13, 15, 21, 35, 39, 45, 63, 65, 91, 105, 117, 195, 273, 315, 455, 585, 819, 1365, 4095

There is a total of 24 positive divisors.
The sum of these divisors is 8736.
The arithmetic mean is 364.

24 odd divisors

1, 3, 5, 7, 9, 13, 15, 21, 35, 39, 45, 63, 65, 91, 105, 117, 195, 273, 315, 455, 585, 819, 1365, 4095

How to compute the divisors of 4095?

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 4095 by each of the numbers from 1 to 4095 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)

4095 / 1 = 4095 (the remainder is 0, so 1 is a divisor of 4095)
4095 / 2 = 2047.5 (the remainder is 1, so 2 is not a divisor of 4095)
4095 / 3 = 1365 (the remainder is 0, so 3 is a divisor of 4095)
...
4095 / 4094 = 1.0002442598925 (the remainder is 1, so 4094 is not a divisor of 4095)
4095 / 4095 = 1 (the remainder is 0, so 4095 is a divisor of 4095)

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 4095 (i.e. 63.992187023105). 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 $\frac{N}{d} = D$ )

4095 / 1 = 4095 (the remainder is 0, so 1 and 4095 are divisors of 4095)
4095 / 2 = 2047.5 (the remainder is 1, so 2 is not a divisor of 4095)
4095 / 3 = 1365 (the remainder is 0, so 3 and 1365 are divisors of 4095)
...
4095 / 62 = 66.048387096774 (the remainder is 3, so 62 is not a divisor of 4095)
4095 / 63 = 65 (the remainder is 0, so 63 and 65 are divisors of 4095)