What are the divisors of 1710?

1, 2, 3, 5, 6, 9, 10, 15, 18, 19, 30, 38, 45, 57, 90, 95, 114, 171, 190, 285, 342, 570, 855, 1710

There is a total of 24 positive divisors.
The sum of these divisors is 4680.
The arithmetic mean is 195.

12 even divisors

2, 6, 10, 18, 30, 38, 90, 114, 190, 342, 570, 1710

12 odd divisors

1, 3, 5, 9, 15, 19, 45, 57, 95, 171, 285, 855

How to compute the divisors of 1710?

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 1710 by each of the numbers from 1 to 1710 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)

1710 / 1 = 1710 (the remainder is 0, so 1 is a divisor of 1710)
1710 / 2 = 855 (the remainder is 0, so 2 is a divisor of 1710)
1710 / 3 = 570 (the remainder is 0, so 3 is a divisor of 1710)
...
1710 / 1709 = 1.0005851375073 (the remainder is 1, so 1709 is not a divisor of 1710)
1710 / 1710 = 1 (the remainder is 0, so 1710 is a divisor of 1710)

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 1710 (i.e. 41.352146256271). 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 $\frac{N}{d} = D$ )

1710 / 1 = 1710 (the remainder is 0, so 1 and 1710 are divisors of 1710)
1710 / 2 = 855 (the remainder is 0, so 2 and 855 are divisors of 1710)
1710 / 3 = 570 (the remainder is 0, so 3 and 570 are divisors of 1710)
...
1710 / 40 = 42.75 (the remainder is 30, so 40 is not a divisor of 1710)
1710 / 41 = 41.707317073171 (the remainder is 29, so 41 is not a divisor of 1710)