What are the divisors of 1767?

1, 3, 19, 31, 57, 93, 589, 1767

There is a total of 8 positive divisors.
The sum of these divisors is 2560.
The arithmetic mean is 320.

8 odd divisors

1, 3, 19, 31, 57, 93, 589, 1767

How to compute the divisors of 1767?

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 1767 by each of the numbers from 1 to 1767 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)

1767 / 1 = 1767 (the remainder is 0, so 1 is a divisor of 1767)
1767 / 2 = 883.5 (the remainder is 1, so 2 is not a divisor of 1767)
1767 / 3 = 589 (the remainder is 0, so 3 is a divisor of 1767)
...
1767 / 1766 = 1.0005662514156 (the remainder is 1, so 1766 is not a divisor of 1767)
1767 / 1767 = 1 (the remainder is 0, so 1767 is a divisor of 1767)

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 1767 (i.e. 42.035699113967). 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 $\frac{N}{d} = D$ )

1767 / 1 = 1767 (the remainder is 0, so 1 and 1767 are divisors of 1767)
1767 / 2 = 883.5 (the remainder is 1, so 2 is not a divisor of 1767)
1767 / 3 = 589 (the remainder is 0, so 3 and 589 are divisors of 1767)
...
1767 / 41 = 43.09756097561 (the remainder is 4, so 41 is not a divisor of 1767)
1767 / 42 = 42.071428571429 (the remainder is 3, so 42 is not a divisor of 1767)