What are the divisors of 1706?

1, 2, 853, 1706

There is a total of 4 positive divisors.
The sum of these divisors is 2562.
The arithmetic mean is 640.5.

2 even divisors

2, 1706

2 odd divisors

1, 853

How to compute the divisors of 1706?

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 1706 by each of the numbers from 1 to 1706 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)

1706 / 1 = 1706 (the remainder is 0, so 1 is a divisor of 1706)
1706 / 2 = 853 (the remainder is 0, so 2 is a divisor of 1706)
1706 / 3 = 568.66666666667 (the remainder is 2, so 3 is not a divisor of 1706)
...
1706 / 1705 = 1.0005865102639 (the remainder is 1, so 1705 is not a divisor of 1706)
1706 / 1706 = 1 (the remainder is 0, so 1706 is a divisor of 1706)

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 1706 (i.e. 41.303752856127). 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 $\frac{N}{d} = D$ )

1706 / 1 = 1706 (the remainder is 0, so 1 and 1706 are divisors of 1706)
1706 / 2 = 853 (the remainder is 0, so 2 and 853 are divisors of 1706)
1706 / 3 = 568.66666666667 (the remainder is 2, so 3 is not a divisor of 1706)
...
1706 / 40 = 42.65 (the remainder is 26, so 40 is not a divisor of 1706)
1706 / 41 = 41.609756097561 (the remainder is 25, so 41 is not a divisor of 1706)