What are the divisors of 2436?

1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 29, 42, 58, 84, 87, 116, 174, 203, 348, 406, 609, 812, 1218, 2436

There is a total of 24 positive divisors.
The sum of these divisors is 6720.
The arithmetic mean is 280.

16 even divisors

2, 4, 6, 12, 14, 28, 42, 58, 84, 116, 174, 348, 406, 812, 1218, 2436

8 odd divisors

1, 3, 7, 21, 29, 87, 203, 609

How to compute the divisors of 2436?

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 2436 by each of the numbers from 1 to 2436 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)

2436 / 1 = 2436 (the remainder is 0, so 1 is a divisor of 2436)
2436 / 2 = 1218 (the remainder is 0, so 2 is a divisor of 2436)
2436 / 3 = 812 (the remainder is 0, so 3 is a divisor of 2436)
...
2436 / 2435 = 1.0004106776181 (the remainder is 1, so 2435 is not a divisor of 2436)
2436 / 2436 = 1 (the remainder is 0, so 2436 is a divisor of 2436)

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 2436 (i.e. 49.355850717012). 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 $\frac{N}{d} = D$ )

2436 / 1 = 2436 (the remainder is 0, so 1 and 2436 are divisors of 2436)
2436 / 2 = 1218 (the remainder is 0, so 2 and 1218 are divisors of 2436)
2436 / 3 = 812 (the remainder is 0, so 3 and 812 are divisors of 2436)
...
2436 / 48 = 50.75 (the remainder is 36, so 48 is not a divisor of 2436)
2436 / 49 = 49.714285714286 (the remainder is 35, so 49 is not a divisor of 2436)