What are the divisors of 1740?

1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 29, 30, 58, 60, 87, 116, 145, 174, 290, 348, 435, 580, 870, 1740

There is a total of 24 positive divisors.
The sum of these divisors is 5040.
The arithmetic mean is 210.

16 even divisors

2, 4, 6, 10, 12, 20, 30, 58, 60, 116, 174, 290, 348, 580, 870, 1740

8 odd divisors

1, 3, 5, 15, 29, 87, 145, 435

How to compute the divisors of 1740?

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 1740 by each of the numbers from 1 to 1740 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)

1740 / 1 = 1740 (the remainder is 0, so 1 is a divisor of 1740)
1740 / 2 = 870 (the remainder is 0, so 2 is a divisor of 1740)
1740 / 3 = 580 (the remainder is 0, so 3 is a divisor of 1740)
...
1740 / 1739 = 1.0005750431282 (the remainder is 1, so 1739 is not a divisor of 1740)
1740 / 1740 = 1 (the remainder is 0, so 1740 is a divisor of 1740)

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 1740 (i.e. 41.713307229228). 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 $\frac{N}{d} = D$ )

1740 / 1 = 1740 (the remainder is 0, so 1 and 1740 are divisors of 1740)
1740 / 2 = 870 (the remainder is 0, so 2 and 870 are divisors of 1740)
1740 / 3 = 580 (the remainder is 0, so 3 and 580 are divisors of 1740)
...
1740 / 40 = 43.5 (the remainder is 20, so 40 is not a divisor of 1740)
1740 / 41 = 42.439024390244 (the remainder is 18, so 41 is not a divisor of 1740)