What are the divisors of 2070?

1, 2, 3, 5, 6, 9, 10, 15, 18, 23, 30, 45, 46, 69, 90, 115, 138, 207, 230, 345, 414, 690, 1035, 2070

There is a total of 24 positive divisors.
The sum of these divisors is 5616.
The arithmetic mean is 234.

12 even divisors

2, 6, 10, 18, 30, 46, 90, 138, 230, 414, 690, 2070

12 odd divisors

1, 3, 5, 9, 15, 23, 45, 69, 115, 207, 345, 1035

How to compute the divisors of 2070?

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 2070 by each of the numbers from 1 to 2070 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)

2070 / 1 = 2070 (the remainder is 0, so 1 is a divisor of 2070)
2070 / 2 = 1035 (the remainder is 0, so 2 is a divisor of 2070)
2070 / 3 = 690 (the remainder is 0, so 3 is a divisor of 2070)
...
2070 / 2069 = 1.0004833252779 (the remainder is 1, so 2069 is not a divisor of 2070)
2070 / 2070 = 1 (the remainder is 0, so 2070 is a divisor of 2070)

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 2070 (i.e. 45.497252664309). 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 $\frac{N}{d} = D$ )

2070 / 1 = 2070 (the remainder is 0, so 1 and 2070 are divisors of 2070)
2070 / 2 = 1035 (the remainder is 0, so 2 and 1035 are divisors of 2070)
2070 / 3 = 690 (the remainder is 0, so 3 and 690 are divisors of 2070)
...
2070 / 44 = 47.045454545455 (the remainder is 2, so 44 is not a divisor of 2070)
2070 / 45 = 46 (the remainder is 0, so 45 and 46 are divisors of 2070)