What are the divisors of 5433?

1, 3, 1811, 5433

There is a total of 4 positive divisors.
The sum of these divisors is 7248.
The arithmetic mean is 1812.

4 odd divisors

1, 3, 1811, 5433

How to compute the divisors of 5433?

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 5433 by each of the numbers from 1 to 5433 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)

5433 / 1 = 5433 (the remainder is 0, so 1 is a divisor of 5433)
5433 / 2 = 2716.5 (the remainder is 1, so 2 is not a divisor of 5433)
5433 / 3 = 1811 (the remainder is 0, so 3 is a divisor of 5433)
...
5433 / 5432 = 1.0001840942563 (the remainder is 1, so 5432 is not a divisor of 5433)
5433 / 5433 = 1 (the remainder is 0, so 5433 is a divisor of 5433)

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 5433 (i.e. 73.708886845482). 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 $\frac{N}{d} = D$ )

5433 / 1 = 5433 (the remainder is 0, so 1 and 5433 are divisors of 5433)
5433 / 2 = 2716.5 (the remainder is 1, so 2 is not a divisor of 5433)
5433 / 3 = 1811 (the remainder is 0, so 3 and 1811 are divisors of 5433)
...
5433 / 72 = 75.458333333333 (the remainder is 33, so 72 is not a divisor of 5433)
5433 / 73 = 74.424657534247 (the remainder is 31, so 73 is not a divisor of 5433)