What are the divisors of 2529?

1, 3, 9, 281, 843, 2529

There is a total of 6 positive divisors.
The sum of these divisors is 3666.
The arithmetic mean is 611.

6 odd divisors

1, 3, 9, 281, 843, 2529

How to compute the divisors of 2529?

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 2529 by each of the numbers from 1 to 2529 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)

2529 / 1 = 2529 (the remainder is 0, so 1 is a divisor of 2529)
2529 / 2 = 1264.5 (the remainder is 1, so 2 is not a divisor of 2529)
2529 / 3 = 843 (the remainder is 0, so 3 is a divisor of 2529)
...
2529 / 2528 = 1.0003955696203 (the remainder is 1, so 2528 is not a divisor of 2529)
2529 / 2529 = 1 (the remainder is 0, so 2529 is a divisor of 2529)

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 2529 (i.e. 50.289163842721). 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 $\frac{N}{d} = D$ )

2529 / 1 = 2529 (the remainder is 0, so 1 and 2529 are divisors of 2529)
2529 / 2 = 1264.5 (the remainder is 1, so 2 is not a divisor of 2529)
2529 / 3 = 843 (the remainder is 0, so 3 and 843 are divisors of 2529)
...
2529 / 49 = 51.612244897959 (the remainder is 30, so 49 is not a divisor of 2529)
2529 / 50 = 50.58 (the remainder is 29, so 50 is not a divisor of 2529)