What are the divisors of 2480?

1, 2, 4, 5, 8, 10, 16, 20, 31, 40, 62, 80, 124, 155, 248, 310, 496, 620, 1240, 2480

There is a total of 20 positive divisors.
The sum of these divisors is 5952.
The arithmetic mean is 297.6.

16 even divisors

2, 4, 8, 10, 16, 20, 40, 62, 80, 124, 248, 310, 496, 620, 1240, 2480

4 odd divisors

1, 5, 31, 155

How to compute the divisors of 2480?

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 2480 by each of the numbers from 1 to 2480 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)

2480 / 1 = 2480 (the remainder is 0, so 1 is a divisor of 2480)
2480 / 2 = 1240 (the remainder is 0, so 2 is a divisor of 2480)
2480 / 3 = 826.66666666667 (the remainder is 2, so 3 is not a divisor of 2480)
...
2480 / 2479 = 1.0004033884631 (the remainder is 1, so 2479 is not a divisor of 2480)
2480 / 2480 = 1 (the remainder is 0, so 2480 is a divisor of 2480)

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 2480 (i.e. 49.799598391955). 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 $\frac{N}{d} = D$ )

2480 / 1 = 2480 (the remainder is 0, so 1 and 2480 are divisors of 2480)
2480 / 2 = 1240 (the remainder is 0, so 2 and 1240 are divisors of 2480)
2480 / 3 = 826.66666666667 (the remainder is 2, so 3 is not a divisor of 2480)
...
2480 / 48 = 51.666666666667 (the remainder is 32, so 48 is not a divisor of 2480)
2480 / 49 = 50.612244897959 (the remainder is 30, so 49 is not a divisor of 2480)