What are the divisors of 3045?

1, 3, 5, 7, 15, 21, 29, 35, 87, 105, 145, 203, 435, 609, 1015, 3045

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, 7, 15, 21, 29, 35, 87, 105, 145, 203, 435, 609, 1015, 3045

How to compute the divisors of 3045?

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 3045 by each of the numbers from 1 to 3045 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)

3045 / 1 = 3045 (the remainder is 0, so 1 is a divisor of 3045)
3045 / 2 = 1522.5 (the remainder is 1, so 2 is not a divisor of 3045)
3045 / 3 = 1015 (the remainder is 0, so 3 is a divisor of 3045)
...
3045 / 3044 = 1.0003285151117 (the remainder is 1, so 3044 is not a divisor of 3045)
3045 / 3045 = 1 (the remainder is 0, so 3045 is a divisor of 3045)

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 3045 (i.e. 55.181518645286). 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 $\frac{N}{d} = D$ )

3045 / 1 = 3045 (the remainder is 0, so 1 and 3045 are divisors of 3045)
3045 / 2 = 1522.5 (the remainder is 1, so 2 is not a divisor of 3045)
3045 / 3 = 1015 (the remainder is 0, so 3 and 1015 are divisors of 3045)
...
3045 / 54 = 56.388888888889 (the remainder is 21, so 54 is not a divisor of 3045)
3045 / 55 = 55.363636363636 (the remainder is 20, so 55 is not a divisor of 3045)