What are the divisors of 2442?

1, 2, 3, 6, 11, 22, 33, 37, 66, 74, 111, 222, 407, 814, 1221, 2442

There is a total of 16 positive divisors.
The sum of these divisors is 5472.
The arithmetic mean is 342.

8 even divisors

2, 6, 22, 66, 74, 222, 814, 2442

8 odd divisors

1, 3, 11, 33, 37, 111, 407, 1221

How to compute the divisors of 2442?

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 2442 by each of the numbers from 1 to 2442 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)

2442 / 1 = 2442 (the remainder is 0, so 1 is a divisor of 2442)
2442 / 2 = 1221 (the remainder is 0, so 2 is a divisor of 2442)
2442 / 3 = 814 (the remainder is 0, so 3 is a divisor of 2442)
...
2442 / 2441 = 1.0004096681688 (the remainder is 1, so 2441 is not a divisor of 2442)
2442 / 2442 = 1 (the remainder is 0, so 2442 is a divisor of 2442)

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 2442 (i.e. 49.416596402423). 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 $\frac{N}{d} = D$ )

2442 / 1 = 2442 (the remainder is 0, so 1 and 2442 are divisors of 2442)
2442 / 2 = 1221 (the remainder is 0, so 2 and 1221 are divisors of 2442)
2442 / 3 = 814 (the remainder is 0, so 3 and 814 are divisors of 2442)
...
2442 / 48 = 50.875 (the remainder is 42, so 48 is not a divisor of 2442)
2442 / 49 = 49.836734693878 (the remainder is 41, so 49 is not a divisor of 2442)