What are the divisors of 3519?

1, 3, 9, 17, 23, 51, 69, 153, 207, 391, 1173, 3519

There is a total of 12 positive divisors.
The sum of these divisors is 5616.
The arithmetic mean is 468.

12 odd divisors

1, 3, 9, 17, 23, 51, 69, 153, 207, 391, 1173, 3519

How to compute the divisors of 3519?

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 3519 by each of the numbers from 1 to 3519 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)

3519 / 1 = 3519 (the remainder is 0, so 1 is a divisor of 3519)
3519 / 2 = 1759.5 (the remainder is 1, so 2 is not a divisor of 3519)
3519 / 3 = 1173 (the remainder is 0, so 3 is a divisor of 3519)
...
3519 / 3518 = 1.0002842524161 (the remainder is 1, so 3518 is not a divisor of 3519)
3519 / 3519 = 1 (the remainder is 0, so 3519 is a divisor of 3519)

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 3519 (i.e. 59.321159799856). 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 $\frac{N}{d} = D$ )

3519 / 1 = 3519 (the remainder is 0, so 1 and 3519 are divisors of 3519)
3519 / 2 = 1759.5 (the remainder is 1, so 2 is not a divisor of 3519)
3519 / 3 = 1173 (the remainder is 0, so 3 and 1173 are divisors of 3519)
...
3519 / 58 = 60.672413793103 (the remainder is 39, so 58 is not a divisor of 3519)
3519 / 59 = 59.64406779661 (the remainder is 38, so 59 is not a divisor of 3519)