What are the divisors of 1758?

1, 2, 3, 6, 293, 586, 879, 1758

There is a total of 8 positive divisors.
The sum of these divisors is 3528.
The arithmetic mean is 441.

4 even divisors

2, 6, 586, 1758

4 odd divisors

1, 3, 293, 879

How to compute the divisors of 1758?

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 1758 by each of the numbers from 1 to 1758 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)

1758 / 1 = 1758 (the remainder is 0, so 1 is a divisor of 1758)
1758 / 2 = 879 (the remainder is 0, so 2 is a divisor of 1758)
1758 / 3 = 586 (the remainder is 0, so 3 is a divisor of 1758)
...
1758 / 1757 = 1.0005691519636 (the remainder is 1, so 1757 is not a divisor of 1758)
1758 / 1758 = 1 (the remainder is 0, so 1758 is a divisor of 1758)

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 1758 (i.e. 41.928510586473). 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 $\frac{N}{d} = D$ )

1758 / 1 = 1758 (the remainder is 0, so 1 and 1758 are divisors of 1758)
1758 / 2 = 879 (the remainder is 0, so 2 and 879 are divisors of 1758)
1758 / 3 = 586 (the remainder is 0, so 3 and 586 are divisors of 1758)
...
1758 / 40 = 43.95 (the remainder is 38, so 40 is not a divisor of 1758)
1758 / 41 = 42.878048780488 (the remainder is 36, so 41 is not a divisor of 1758)