What are the divisors of 2412?

1, 2, 3, 4, 6, 9, 12, 18, 36, 67, 134, 201, 268, 402, 603, 804, 1206, 2412

There is a total of 18 positive divisors.
The sum of these divisors is 6188.
The arithmetic mean is 343.77777777778.

12 even divisors

2, 4, 6, 12, 18, 36, 134, 268, 402, 804, 1206, 2412

6 odd divisors

1, 3, 9, 67, 201, 603

How to compute the divisors of 2412?

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 2412 by each of the numbers from 1 to 2412 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)

2412 / 1 = 2412 (the remainder is 0, so 1 is a divisor of 2412)
2412 / 2 = 1206 (the remainder is 0, so 2 is a divisor of 2412)
2412 / 3 = 804 (the remainder is 0, so 3 is a divisor of 2412)
...
2412 / 2411 = 1.0004147656574 (the remainder is 1, so 2411 is not a divisor of 2412)
2412 / 2412 = 1 (the remainder is 0, so 2412 is a divisor of 2412)

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 2412 (i.e. 49.112116631235). 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 $\frac{N}{d} = D$ )

2412 / 1 = 2412 (the remainder is 0, so 1 and 2412 are divisors of 2412)
2412 / 2 = 1206 (the remainder is 0, so 2 and 1206 are divisors of 2412)
2412 / 3 = 804 (the remainder is 0, so 3 and 804 are divisors of 2412)
...
2412 / 48 = 50.25 (the remainder is 12, so 48 is not a divisor of 2412)
2412 / 49 = 49.224489795918 (the remainder is 11, so 49 is not a divisor of 2412)