What are the divisors of 5103?

1, 3, 7, 9, 21, 27, 63, 81, 189, 243, 567, 729, 1701, 5103

There is a total of 14 positive divisors.
The sum of these divisors is 8744.
The arithmetic mean is 624.57142857143.

14 odd divisors

1, 3, 7, 9, 21, 27, 63, 81, 189, 243, 567, 729, 1701, 5103

How to compute the divisors of 5103?

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 5103 by each of the numbers from 1 to 5103 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)

5103 / 1 = 5103 (the remainder is 0, so 1 is a divisor of 5103)
5103 / 2 = 2551.5 (the remainder is 1, so 2 is not a divisor of 5103)
5103 / 3 = 1701 (the remainder is 0, so 3 is a divisor of 5103)
...
5103 / 5102 = 1.000196001568 (the remainder is 1, so 5102 is not a divisor of 5103)
5103 / 5103 = 1 (the remainder is 0, so 5103 is a divisor of 5103)

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 5103 (i.e. 71.435285398744). 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 $\frac{N}{d} = D$ )

5103 / 1 = 5103 (the remainder is 0, so 1 and 5103 are divisors of 5103)
5103 / 2 = 2551.5 (the remainder is 1, so 2 is not a divisor of 5103)
5103 / 3 = 1701 (the remainder is 0, so 3 and 1701 are divisors of 5103)
...
5103 / 70 = 72.9 (the remainder is 63, so 70 is not a divisor of 5103)
5103 / 71 = 71.87323943662 (the remainder is 62, so 71 is not a divisor of 5103)