What are the divisors of 1797?

1, 3, 599, 1797

There is a total of 4 positive divisors.
The sum of these divisors is 2400.
The arithmetic mean is 600.

4 odd divisors

1, 3, 599, 1797

How to compute the divisors of 1797?

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 1797 by each of the numbers from 1 to 1797 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)

1797 / 1 = 1797 (the remainder is 0, so 1 is a divisor of 1797)
1797 / 2 = 898.5 (the remainder is 1, so 2 is not a divisor of 1797)
1797 / 3 = 599 (the remainder is 0, so 3 is a divisor of 1797)
...
1797 / 1796 = 1.0005567928731 (the remainder is 1, so 1796 is not a divisor of 1797)
1797 / 1797 = 1 (the remainder is 0, so 1797 is a divisor of 1797)

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 1797 (i.e. 42.391036788453). 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 $\frac{N}{d} = D$ )

1797 / 1 = 1797 (the remainder is 0, so 1 and 1797 are divisors of 1797)
1797 / 2 = 898.5 (the remainder is 1, so 2 is not a divisor of 1797)
1797 / 3 = 599 (the remainder is 0, so 3 and 599 are divisors of 1797)
...
1797 / 41 = 43.829268292683 (the remainder is 34, so 41 is not a divisor of 1797)
1797 / 42 = 42.785714285714 (the remainder is 33, so 42 is not a divisor of 1797)