What are the divisors of 3270?

1, 2, 3, 5, 6, 10, 15, 30, 109, 218, 327, 545, 654, 1090, 1635, 3270

There is a total of 16 positive divisors.
The sum of these divisors is 7920.
The arithmetic mean is 495.

8 even divisors

2, 6, 10, 30, 218, 654, 1090, 3270

8 odd divisors

1, 3, 5, 15, 109, 327, 545, 1635

How to compute the divisors of 3270?

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 3270 by each of the numbers from 1 to 3270 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)

3270 / 1 = 3270 (the remainder is 0, so 1 is a divisor of 3270)
3270 / 2 = 1635 (the remainder is 0, so 2 is a divisor of 3270)
3270 / 3 = 1090 (the remainder is 0, so 3 is a divisor of 3270)
...
3270 / 3269 = 1.0003059039462 (the remainder is 1, so 3269 is not a divisor of 3270)
3270 / 3270 = 1 (the remainder is 0, so 3270 is a divisor of 3270)

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 3270 (i.e. 57.183913821983). 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 $\frac{N}{d} = D$ )

3270 / 1 = 3270 (the remainder is 0, so 1 and 3270 are divisors of 3270)
3270 / 2 = 1635 (the remainder is 0, so 2 and 1635 are divisors of 3270)
3270 / 3 = 1090 (the remainder is 0, so 3 and 1090 are divisors of 3270)
...
3270 / 56 = 58.392857142857 (the remainder is 22, so 56 is not a divisor of 3270)
3270 / 57 = 57.368421052632 (the remainder is 21, so 57 is not a divisor of 3270)