What are the divisors of 1705?

1, 5, 11, 31, 55, 155, 341, 1705

There is a total of 8 positive divisors.
The sum of these divisors is 2304.
The arithmetic mean is 288.

8 odd divisors

1, 5, 11, 31, 55, 155, 341, 1705

How to compute the divisors of 1705?

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 1705 by each of the numbers from 1 to 1705 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)

1705 / 1 = 1705 (the remainder is 0, so 1 is a divisor of 1705)
1705 / 2 = 852.5 (the remainder is 1, so 2 is not a divisor of 1705)
1705 / 3 = 568.33333333333 (the remainder is 1, so 3 is not a divisor of 1705)
...
1705 / 1704 = 1.0005868544601 (the remainder is 1, so 1704 is not a divisor of 1705)
1705 / 1705 = 1 (the remainder is 0, so 1705 is a divisor of 1705)

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 1705 (i.e. 41.291645644125). 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 $\frac{N}{d} = D$ )

1705 / 1 = 1705 (the remainder is 0, so 1 and 1705 are divisors of 1705)
1705 / 2 = 852.5 (the remainder is 1, so 2 is not a divisor of 1705)
1705 / 3 = 568.33333333333 (the remainder is 1, so 3 is not a divisor of 1705)
...
1705 / 40 = 42.625 (the remainder is 25, so 40 is not a divisor of 1705)
1705 / 41 = 41.585365853659 (the remainder is 24, so 41 is not a divisor of 1705)