What are the divisors of 2420?

1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110, 121, 220, 242, 484, 605, 1210, 2420

There is a total of 18 positive divisors.
The sum of these divisors is 5586.
The arithmetic mean is 310.33333333333.

12 even divisors

2, 4, 10, 20, 22, 44, 110, 220, 242, 484, 1210, 2420

6 odd divisors

1, 5, 11, 55, 121, 605

How to compute the divisors of 2420?

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 2420 by each of the numbers from 1 to 2420 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)

2420 / 1 = 2420 (the remainder is 0, so 1 is a divisor of 2420)
2420 / 2 = 1210 (the remainder is 0, so 2 is a divisor of 2420)
2420 / 3 = 806.66666666667 (the remainder is 2, so 3 is not a divisor of 2420)
...
2420 / 2419 = 1.0004133939644 (the remainder is 1, so 2419 is not a divisor of 2420)
2420 / 2420 = 1 (the remainder is 0, so 2420 is a divisor of 2420)

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 2420 (i.e. 49.193495504995). 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 $\frac{N}{d} = D$ )

2420 / 1 = 2420 (the remainder is 0, so 1 and 2420 are divisors of 2420)
2420 / 2 = 1210 (the remainder is 0, so 2 and 1210 are divisors of 2420)
2420 / 3 = 806.66666666667 (the remainder is 2, so 3 is not a divisor of 2420)
...
2420 / 48 = 50.416666666667 (the remainder is 20, so 48 is not a divisor of 2420)
2420 / 49 = 49.387755102041 (the remainder is 19, so 49 is not a divisor of 2420)