What are the divisors of 1734?

1, 2, 3, 6, 17, 34, 51, 102, 289, 578, 867, 1734

There is a total of 12 positive divisors.
The sum of these divisors is 3684.
The arithmetic mean is 307.

6 even divisors

2, 6, 34, 102, 578, 1734

6 odd divisors

1, 3, 17, 51, 289, 867

How to compute the divisors of 1734?

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 1734 by each of the numbers from 1 to 1734 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)

1734 / 1 = 1734 (the remainder is 0, so 1 is a divisor of 1734)
1734 / 2 = 867 (the remainder is 0, so 2 is a divisor of 1734)
1734 / 3 = 578 (the remainder is 0, so 3 is a divisor of 1734)
...
1734 / 1733 = 1.000577034045 (the remainder is 1, so 1733 is not a divisor of 1734)
1734 / 1734 = 1 (the remainder is 0, so 1734 is a divisor of 1734)

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 1734 (i.e. 41.641325627314). 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 $\frac{N}{d} = D$ )

1734 / 1 = 1734 (the remainder is 0, so 1 and 1734 are divisors of 1734)
1734 / 2 = 867 (the remainder is 0, so 2 and 867 are divisors of 1734)
1734 / 3 = 578 (the remainder is 0, so 3 and 578 are divisors of 1734)
...
1734 / 40 = 43.35 (the remainder is 14, so 40 is not a divisor of 1734)
1734 / 41 = 42.292682926829 (the remainder is 12, so 41 is not a divisor of 1734)