What are the divisors of 3795?

1, 3, 5, 11, 15, 23, 33, 55, 69, 115, 165, 253, 345, 759, 1265, 3795

There is a total of 16 positive divisors.
The sum of these divisors is 6912.
The arithmetic mean is 432.

16 odd divisors

1, 3, 5, 11, 15, 23, 33, 55, 69, 115, 165, 253, 345, 759, 1265, 3795

How to compute the divisors of 3795?

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 3795 by each of the numbers from 1 to 3795 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)

3795 / 1 = 3795 (the remainder is 0, so 1 is a divisor of 3795)
3795 / 2 = 1897.5 (the remainder is 1, so 2 is not a divisor of 3795)
3795 / 3 = 1265 (the remainder is 0, so 3 is a divisor of 3795)
...
3795 / 3794 = 1.0002635740643 (the remainder is 1, so 3794 is not a divisor of 3795)
3795 / 3795 = 1 (the remainder is 0, so 3795 is a divisor of 3795)

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 3795 (i.e. 61.603571325046). 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 $\frac{N}{d} = D$ )

3795 / 1 = 3795 (the remainder is 0, so 1 and 3795 are divisors of 3795)
3795 / 2 = 1897.5 (the remainder is 1, so 2 is not a divisor of 3795)
3795 / 3 = 1265 (the remainder is 0, so 3 and 1265 are divisors of 3795)
...
3795 / 60 = 63.25 (the remainder is 15, so 60 is not a divisor of 3795)
3795 / 61 = 62.213114754098 (the remainder is 13, so 61 is not a divisor of 3795)