What are the divisors of 1737?

1, 3, 9, 193, 579, 1737

There is a total of 6 positive divisors.
The sum of these divisors is 2522.
The arithmetic mean is 420.33333333333.

6 odd divisors

1, 3, 9, 193, 579, 1737

How to compute the divisors of 1737?

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 1737 by each of the numbers from 1 to 1737 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)

1737 / 1 = 1737 (the remainder is 0, so 1 is a divisor of 1737)
1737 / 2 = 868.5 (the remainder is 1, so 2 is not a divisor of 1737)
1737 / 3 = 579 (the remainder is 0, so 3 is a divisor of 1737)
...
1737 / 1736 = 1.0005760368664 (the remainder is 1, so 1736 is not a divisor of 1737)
1737 / 1737 = 1 (the remainder is 0, so 1737 is a divisor of 1737)

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 1737 (i.e. 41.677331968349). 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 $\frac{N}{d} = D$ )

1737 / 1 = 1737 (the remainder is 0, so 1 and 1737 are divisors of 1737)
1737 / 2 = 868.5 (the remainder is 1, so 2 is not a divisor of 1737)
1737 / 3 = 579 (the remainder is 0, so 3 and 579 are divisors of 1737)
...
1737 / 40 = 43.425 (the remainder is 17, so 40 is not a divisor of 1737)
1737 / 41 = 42.365853658537 (the remainder is 15, so 41 is not a divisor of 1737)