What are the divisors of 1773?

1, 3, 9, 197, 591, 1773

There is a total of 6 positive divisors.
The sum of these divisors is 2574.
The arithmetic mean is 429.

6 odd divisors

1, 3, 9, 197, 591, 1773

How to compute the divisors of 1773?

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 1773 by each of the numbers from 1 to 1773 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)

1773 / 1 = 1773 (the remainder is 0, so 1 is a divisor of 1773)
1773 / 2 = 886.5 (the remainder is 1, so 2 is not a divisor of 1773)
1773 / 3 = 591 (the remainder is 0, so 3 is a divisor of 1773)
...
1773 / 1772 = 1.0005643340858 (the remainder is 1, so 1772 is not a divisor of 1773)
1773 / 1773 = 1 (the remainder is 0, so 1773 is a divisor of 1773)

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 1773 (i.e. 42.107006542855). 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 $\frac{N}{d} = D$ )

1773 / 1 = 1773 (the remainder is 0, so 1 and 1773 are divisors of 1773)
1773 / 2 = 886.5 (the remainder is 1, so 2 is not a divisor of 1773)
1773 / 3 = 591 (the remainder is 0, so 3 and 591 are divisors of 1773)
...
1773 / 41 = 43.243902439024 (the remainder is 10, so 41 is not a divisor of 1773)
1773 / 42 = 42.214285714286 (the remainder is 9, so 42 is not a divisor of 1773)