What are the divisors of 3015?

1, 3, 5, 9, 15, 45, 67, 201, 335, 603, 1005, 3015

There is a total of 12 positive divisors.
The sum of these divisors is 5304.
The arithmetic mean is 442.

12 odd divisors

1, 3, 5, 9, 15, 45, 67, 201, 335, 603, 1005, 3015

How to compute the divisors of 3015?

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 3015 by each of the numbers from 1 to 3015 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)

3015 / 1 = 3015 (the remainder is 0, so 1 is a divisor of 3015)
3015 / 2 = 1507.5 (the remainder is 1, so 2 is not a divisor of 3015)
3015 / 3 = 1005 (the remainder is 0, so 3 is a divisor of 3015)
...
3015 / 3014 = 1.0003317850033 (the remainder is 1, so 3014 is not a divisor of 3015)
3015 / 3015 = 1 (the remainder is 0, so 3015 is a divisor of 3015)

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 3015 (i.e. 54.909015653169). 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 $\frac{N}{d} = D$ )

3015 / 1 = 3015 (the remainder is 0, so 1 and 3015 are divisors of 3015)
3015 / 2 = 1507.5 (the remainder is 1, so 2 is not a divisor of 3015)
3015 / 3 = 1005 (the remainder is 0, so 3 and 1005 are divisors of 3015)
...
3015 / 53 = 56.88679245283 (the remainder is 47, so 53 is not a divisor of 3015)
3015 / 54 = 55.833333333333 (the remainder is 45, so 54 is not a divisor of 3015)