What are the divisors of 1791?

1, 3, 9, 199, 597, 1791

There is a total of 6 positive divisors.
The sum of these divisors is 2600.
The arithmetic mean is 433.33333333333.

6 odd divisors

1, 3, 9, 199, 597, 1791

How to compute the divisors of 1791?

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 1791 by each of the numbers from 1 to 1791 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)

1791 / 1 = 1791 (the remainder is 0, so 1 is a divisor of 1791)
1791 / 2 = 895.5 (the remainder is 1, so 2 is not a divisor of 1791)
1791 / 3 = 597 (the remainder is 0, so 3 is a divisor of 1791)
...
1791 / 1790 = 1.0005586592179 (the remainder is 1, so 1790 is not a divisor of 1791)
1791 / 1791 = 1 (the remainder is 0, so 1791 is a divisor of 1791)

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 1791 (i.e. 42.320207938998). 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 $\frac{N}{d} = D$ )

1791 / 1 = 1791 (the remainder is 0, so 1 and 1791 are divisors of 1791)
1791 / 2 = 895.5 (the remainder is 1, so 2 is not a divisor of 1791)
1791 / 3 = 597 (the remainder is 0, so 3 and 597 are divisors of 1791)
...
1791 / 41 = 43.682926829268 (the remainder is 28, so 41 is not a divisor of 1791)
1791 / 42 = 42.642857142857 (the remainder is 27, so 42 is not a divisor of 1791)