What are the divisors of 543?

1, 3, 181, 543

There is a total of 4 positive divisors.
The sum of these divisors is 728.
The arithmetic mean is 182.

4 odd divisors

1, 3, 181, 543

How to compute the divisors of 543?

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 543 by each of the numbers from 1 to 543 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)

543 / 1 = 543 (the remainder is 0, so 1 is a divisor of 543)
543 / 2 = 271.5 (the remainder is 1, so 2 is not a divisor of 543)
543 / 3 = 181 (the remainder is 0, so 3 is a divisor of 543)
...
543 / 542 = 1.0018450184502 (the remainder is 1, so 542 is not a divisor of 543)
543 / 543 = 1 (the remainder is 0, so 543 is a divisor of 543)

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 543 (i.e. 23.302360395462). 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 $\frac{N}{d} = D$ )

543 / 1 = 543 (the remainder is 0, so 1 and 543 are divisors of 543)
543 / 2 = 271.5 (the remainder is 1, so 2 is not a divisor of 543)
543 / 3 = 181 (the remainder is 0, so 3 and 181 are divisors of 543)
...
543 / 22 = 24.681818181818 (the remainder is 15, so 22 is not a divisor of 543)
543 / 23 = 23.608695652174 (the remainder is 14, so 23 is not a divisor of 543)