What are the divisors of 6030?

1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 67, 90, 134, 201, 335, 402, 603, 670, 1005, 1206, 2010, 3015, 6030

There is a total of 24 positive divisors.
The sum of these divisors is 15912.
The arithmetic mean is 663.

12 even divisors

2, 6, 10, 18, 30, 90, 134, 402, 670, 1206, 2010, 6030

12 odd divisors

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

How to compute the divisors of 6030?

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 6030 by each of the numbers from 1 to 6030 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)

6030 / 1 = 6030 (the remainder is 0, so 1 is a divisor of 6030)
6030 / 2 = 3015 (the remainder is 0, so 2 is a divisor of 6030)
6030 / 3 = 2010 (the remainder is 0, so 3 is a divisor of 6030)
...
6030 / 6029 = 1.0001658649859 (the remainder is 1, so 6029 is not a divisor of 6030)
6030 / 6030 = 1 (the remainder is 0, so 6030 is a divisor of 6030)

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 6030 (i.e. 77.653074633269). 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 $\frac{N}{d} = D$ )

6030 / 1 = 6030 (the remainder is 0, so 1 and 6030 are divisors of 6030)
6030 / 2 = 3015 (the remainder is 0, so 2 and 3015 are divisors of 6030)
6030 / 3 = 2010 (the remainder is 0, so 3 and 2010 are divisors of 6030)
...
6030 / 76 = 79.342105263158 (the remainder is 26, so 76 is not a divisor of 6030)
6030 / 77 = 78.311688311688 (the remainder is 24, so 77 is not a divisor of 6030)