What are the divisors of 3207?

1, 3, 1069, 3207

There is a total of 4 positive divisors.
The sum of these divisors is 4280.
The arithmetic mean is 1070.

4 odd divisors

1, 3, 1069, 3207

How to compute the divisors of 3207?

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 3207 by each of the numbers from 1 to 3207 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)

3207 / 1 = 3207 (the remainder is 0, so 1 is a divisor of 3207)
3207 / 2 = 1603.5 (the remainder is 1, so 2 is not a divisor of 3207)
3207 / 3 = 1069 (the remainder is 0, so 3 is a divisor of 3207)
...
3207 / 3206 = 1.0003119151591 (the remainder is 1, so 3206 is not a divisor of 3207)
3207 / 3207 = 1 (the remainder is 0, so 3207 is a divisor of 3207)

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 3207 (i.e. 56.630380539071). 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 $\frac{N}{d} = D$ )

3207 / 1 = 3207 (the remainder is 0, so 1 and 3207 are divisors of 3207)
3207 / 2 = 1603.5 (the remainder is 1, so 2 is not a divisor of 3207)
3207 / 3 = 1069 (the remainder is 0, so 3 and 1069 are divisors of 3207)
...
3207 / 55 = 58.309090909091 (the remainder is 17, so 55 is not a divisor of 3207)
3207 / 56 = 57.267857142857 (the remainder is 15, so 56 is not a divisor of 3207)