What are the divisors of 3210?

1, 2, 3, 5, 6, 10, 15, 30, 107, 214, 321, 535, 642, 1070, 1605, 3210

There is a total of 16 positive divisors.
The sum of these divisors is 7776.
The arithmetic mean is 486.

8 even divisors

2, 6, 10, 30, 214, 642, 1070, 3210

8 odd divisors

1, 3, 5, 15, 107, 321, 535, 1605

How to compute the divisors of 3210?

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 3210 by each of the numbers from 1 to 3210 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)

3210 / 1 = 3210 (the remainder is 0, so 1 is a divisor of 3210)
3210 / 2 = 1605 (the remainder is 0, so 2 is a divisor of 3210)
3210 / 3 = 1070 (the remainder is 0, so 3 is a divisor of 3210)
...
3210 / 3209 = 1.0003116235587 (the remainder is 1, so 3209 is not a divisor of 3210)
3210 / 3210 = 1 (the remainder is 0, so 3210 is a divisor of 3210)

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 3210 (i.e. 56.656861896861). 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 $\frac{N}{d} = D$ )

3210 / 1 = 3210 (the remainder is 0, so 1 and 3210 are divisors of 3210)
3210 / 2 = 1605 (the remainder is 0, so 2 and 1605 are divisors of 3210)
3210 / 3 = 1070 (the remainder is 0, so 3 and 1070 are divisors of 3210)
...
3210 / 55 = 58.363636363636 (the remainder is 20, so 55 is not a divisor of 3210)
3210 / 56 = 57.321428571429 (the remainder is 18, so 56 is not a divisor of 3210)