What are the divisors of 2523?

1, 3, 29, 87, 841, 2523

There is a total of 6 positive divisors.
The sum of these divisors is 3484.
The arithmetic mean is 580.66666666667.

6 odd divisors

1, 3, 29, 87, 841, 2523

How to compute the divisors of 2523?

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 2523 by each of the numbers from 1 to 2523 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)

2523 / 1 = 2523 (the remainder is 0, so 1 is a divisor of 2523)
2523 / 2 = 1261.5 (the remainder is 1, so 2 is not a divisor of 2523)
2523 / 3 = 841 (the remainder is 0, so 3 is a divisor of 2523)
...
2523 / 2522 = 1.0003965107058 (the remainder is 1, so 2522 is not a divisor of 2523)
2523 / 2523 = 1 (the remainder is 0, so 2523 is a divisor of 2523)

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 2523 (i.e. 50.229473419497). 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 $\frac{N}{d} = D$ )

2523 / 1 = 2523 (the remainder is 0, so 1 and 2523 are divisors of 2523)
2523 / 2 = 1261.5 (the remainder is 1, so 2 is not a divisor of 2523)
2523 / 3 = 841 (the remainder is 0, so 3 and 841 are divisors of 2523)
...
2523 / 49 = 51.489795918367 (the remainder is 24, so 49 is not a divisor of 2523)
2523 / 50 = 50.46 (the remainder is 23, so 50 is not a divisor of 2523)