What are the divisors of 5547?

1, 3, 43, 129, 1849, 5547

There is a total of 6 positive divisors.
The sum of these divisors is 7572.
The arithmetic mean is 1262.

6 odd divisors

1, 3, 43, 129, 1849, 5547

How to compute the divisors of 5547?

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 5547 by each of the numbers from 1 to 5547 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)

5547 / 1 = 5547 (the remainder is 0, so 1 is a divisor of 5547)
5547 / 2 = 2773.5 (the remainder is 1, so 2 is not a divisor of 5547)
5547 / 3 = 1849 (the remainder is 0, so 3 is a divisor of 5547)
...
5547 / 5546 = 1.0001803101334 (the remainder is 1, so 5546 is not a divisor of 5547)
5547 / 5547 = 1 (the remainder is 0, so 5547 is a divisor of 5547)

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 5547 (i.e. 74.478184725462). 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 $\frac{N}{d} = D$ )

5547 / 1 = 5547 (the remainder is 0, so 1 and 5547 are divisors of 5547)
5547 / 2 = 2773.5 (the remainder is 1, so 2 is not a divisor of 5547)
5547 / 3 = 1849 (the remainder is 0, so 3 and 1849 are divisors of 5547)
...
5547 / 73 = 75.986301369863 (the remainder is 72, so 73 is not a divisor of 5547)
5547 / 74 = 74.959459459459 (the remainder is 71, so 74 is not a divisor of 5547)