What are the divisors of 3258?

1, 2, 3, 6, 9, 18, 181, 362, 543, 1086, 1629, 3258

There is a total of 12 positive divisors.
The sum of these divisors is 7098.
The arithmetic mean is 591.5.

6 even divisors

2, 6, 18, 362, 1086, 3258

6 odd divisors

1, 3, 9, 181, 543, 1629

How to compute the divisors of 3258?

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 3258 by each of the numbers from 1 to 3258 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)

3258 / 1 = 3258 (the remainder is 0, so 1 is a divisor of 3258)
3258 / 2 = 1629 (the remainder is 0, so 2 is a divisor of 3258)
3258 / 3 = 1086 (the remainder is 0, so 3 is a divisor of 3258)
...
3258 / 3257 = 1.0003070310101 (the remainder is 1, so 3257 is not a divisor of 3258)
3258 / 3258 = 1 (the remainder is 0, so 3258 is a divisor of 3258)

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 3258 (i.e. 57.078892771321). 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 $\frac{N}{d} = D$ )

3258 / 1 = 3258 (the remainder is 0, so 1 and 3258 are divisors of 3258)
3258 / 2 = 1629 (the remainder is 0, so 2 and 1629 are divisors of 3258)
3258 / 3 = 1086 (the remainder is 0, so 3 and 1086 are divisors of 3258)
...
3258 / 56 = 58.178571428571 (the remainder is 10, so 56 is not a divisor of 3258)
3258 / 57 = 57.157894736842 (the remainder is 9, so 57 is not a divisor of 3258)