What are the divisors of 3105?

1, 3, 5, 9, 15, 23, 27, 45, 69, 115, 135, 207, 345, 621, 1035, 3105

There is a total of 16 positive divisors.
The sum of these divisors is 5760.
The arithmetic mean is 360.

16 odd divisors

1, 3, 5, 9, 15, 23, 27, 45, 69, 115, 135, 207, 345, 621, 1035, 3105

How to compute the divisors of 3105?

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 3105 by each of the numbers from 1 to 3105 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)

3105 / 1 = 3105 (the remainder is 0, so 1 is a divisor of 3105)
3105 / 2 = 1552.5 (the remainder is 1, so 2 is not a divisor of 3105)
3105 / 3 = 1035 (the remainder is 0, so 3 is a divisor of 3105)
...
3105 / 3104 = 1.0003221649485 (the remainder is 1, so 3104 is not a divisor of 3105)
3105 / 3105 = 1 (the remainder is 0, so 3105 is a divisor of 3105)

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 3105 (i.e. 55.72252686302). 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 $\frac{N}{d} = D$ )

3105 / 1 = 3105 (the remainder is 0, so 1 and 3105 are divisors of 3105)
3105 / 2 = 1552.5 (the remainder is 1, so 2 is not a divisor of 3105)
3105 / 3 = 1035 (the remainder is 0, so 3 and 1035 are divisors of 3105)
...
3105 / 54 = 57.5 (the remainder is 27, so 54 is not a divisor of 3105)
3105 / 55 = 56.454545454545 (the remainder is 25, so 55 is not a divisor of 3105)