What are the divisors of 3190?

1, 2, 5, 10, 11, 22, 29, 55, 58, 110, 145, 290, 319, 638, 1595, 3190

There is a total of 16 positive divisors.
The sum of these divisors is 6480.
The arithmetic mean is 405.

8 even divisors

2, 10, 22, 58, 110, 290, 638, 3190

8 odd divisors

1, 5, 11, 29, 55, 145, 319, 1595

How to compute the divisors of 3190?

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 3190 by each of the numbers from 1 to 3190 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)

3190 / 1 = 3190 (the remainder is 0, so 1 is a divisor of 3190)
3190 / 2 = 1595 (the remainder is 0, so 2 is a divisor of 3190)
3190 / 3 = 1063.3333333333 (the remainder is 1, so 3 is not a divisor of 3190)
...
3190 / 3189 = 1.0003135779241 (the remainder is 1, so 3189 is not a divisor of 3190)
3190 / 3190 = 1 (the remainder is 0, so 3190 is a divisor of 3190)

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 3190 (i.e. 56.480084985772). 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 $\frac{N}{d} = D$ )

3190 / 1 = 3190 (the remainder is 0, so 1 and 3190 are divisors of 3190)
3190 / 2 = 1595 (the remainder is 0, so 2 and 1595 are divisors of 3190)
3190 / 3 = 1063.3333333333 (the remainder is 1, so 3 is not a divisor of 3190)
...
3190 / 55 = 58 (the remainder is 0, so 55 and 58 are divisors of 3190)
3190 / 56 = 56.964285714286 (the remainder is 54, so 56 is not a divisor of 3190)