What are the divisors of 2440?

1, 2, 4, 5, 8, 10, 20, 40, 61, 122, 244, 305, 488, 610, 1220, 2440

There is a total of 16 positive divisors.
The sum of these divisors is 5580.
The arithmetic mean is 348.75.

12 even divisors

2, 4, 8, 10, 20, 40, 122, 244, 488, 610, 1220, 2440

4 odd divisors

1, 5, 61, 305

How to compute the divisors of 2440?

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 2440 by each of the numbers from 1 to 2440 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)

2440 / 1 = 2440 (the remainder is 0, so 1 is a divisor of 2440)
2440 / 2 = 1220 (the remainder is 0, so 2 is a divisor of 2440)
2440 / 3 = 813.33333333333 (the remainder is 1, so 3 is not a divisor of 2440)
...
2440 / 2439 = 1.0004100041 (the remainder is 1, so 2439 is not a divisor of 2440)
2440 / 2440 = 1 (the remainder is 0, so 2440 is a divisor of 2440)

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 2440 (i.e. 49.396356140914). 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 $\frac{N}{d} = D$ )

2440 / 1 = 2440 (the remainder is 0, so 1 and 2440 are divisors of 2440)
2440 / 2 = 1220 (the remainder is 0, so 2 and 1220 are divisors of 2440)
2440 / 3 = 813.33333333333 (the remainder is 1, so 3 is not a divisor of 2440)
...
2440 / 48 = 50.833333333333 (the remainder is 40, so 48 is not a divisor of 2440)
2440 / 49 = 49.795918367347 (the remainder is 39, so 49 is not a divisor of 2440)