What are the divisors of 1755?

1, 3, 5, 9, 13, 15, 27, 39, 45, 65, 117, 135, 195, 351, 585, 1755

There is a total of 16 positive divisors.
The sum of these divisors is 3360.
The arithmetic mean is 210.

16 odd divisors

1, 3, 5, 9, 13, 15, 27, 39, 45, 65, 117, 135, 195, 351, 585, 1755

How to compute the divisors of 1755?

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 1755 by each of the numbers from 1 to 1755 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)

1755 / 1 = 1755 (the remainder is 0, so 1 is a divisor of 1755)
1755 / 2 = 877.5 (the remainder is 1, so 2 is not a divisor of 1755)
1755 / 3 = 585 (the remainder is 0, so 3 is a divisor of 1755)
...
1755 / 1754 = 1.0005701254276 (the remainder is 1, so 1754 is not a divisor of 1755)
1755 / 1755 = 1 (the remainder is 0, so 1755 is a divisor of 1755)

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 1755 (i.e. 41.892720131307). 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 $\frac{N}{d} = D$ )

1755 / 1 = 1755 (the remainder is 0, so 1 and 1755 are divisors of 1755)
1755 / 2 = 877.5 (the remainder is 1, so 2 is not a divisor of 1755)
1755 / 3 = 585 (the remainder is 0, so 3 and 585 are divisors of 1755)
...
1755 / 40 = 43.875 (the remainder is 35, so 40 is not a divisor of 1755)
1755 / 41 = 42.80487804878 (the remainder is 33, so 41 is not a divisor of 1755)