What are the divisors of 1719?

1, 3, 9, 191, 573, 1719

There is a total of 6 positive divisors.
The sum of these divisors is 2496.
The arithmetic mean is 416.

6 odd divisors

1, 3, 9, 191, 573, 1719

How to compute the divisors of 1719?

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 1719 by each of the numbers from 1 to 1719 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)

1719 / 1 = 1719 (the remainder is 0, so 1 is a divisor of 1719)
1719 / 2 = 859.5 (the remainder is 1, so 2 is not a divisor of 1719)
1719 / 3 = 573 (the remainder is 0, so 3 is a divisor of 1719)
...
1719 / 1718 = 1.0005820721769 (the remainder is 1, so 1718 is not a divisor of 1719)
1719 / 1719 = 1 (the remainder is 0, so 1719 is a divisor of 1719)

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 1719 (i.e. 41.460824883256). 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 $\frac{N}{d} = D$ )

1719 / 1 = 1719 (the remainder is 0, so 1 and 1719 are divisors of 1719)
1719 / 2 = 859.5 (the remainder is 1, so 2 is not a divisor of 1719)
1719 / 3 = 573 (the remainder is 0, so 3 and 573 are divisors of 1719)
...
1719 / 40 = 42.975 (the remainder is 39, so 40 is not a divisor of 1719)
1719 / 41 = 41.926829268293 (the remainder is 38, so 41 is not a divisor of 1719)