What are the divisors of 243?

1, 3, 9, 27, 81, 243

There is a total of 6 positive divisors.
The sum of these divisors is 364.
The arithmetic mean is 60.666666666667.

6 odd divisors

1, 3, 9, 27, 81, 243

How to compute the divisors of 243?

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 243 by each of the numbers from 1 to 243 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)

243 / 1 = 243 (the remainder is 0, so 1 is a divisor of 243)
243 / 2 = 121.5 (the remainder is 1, so 2 is not a divisor of 243)
243 / 3 = 81 (the remainder is 0, so 3 is a divisor of 243)
...
243 / 242 = 1.004132231405 (the remainder is 1, so 242 is not a divisor of 243)
243 / 243 = 1 (the remainder is 0, so 243 is a divisor of 243)

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 243 (i.e. 15.58845726812). 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 $\frac{N}{d} = D$ )

243 / 1 = 243 (the remainder is 0, so 1 and 243 are divisors of 243)
243 / 2 = 121.5 (the remainder is 1, so 2 is not a divisor of 243)
243 / 3 = 81 (the remainder is 0, so 3 and 81 are divisors of 243)
...
243 / 14 = 17.357142857143 (the remainder is 5, so 14 is not a divisor of 243)
243 / 15 = 16.2 (the remainder is 3, so 15 is not a divisor of 243)