What are the divisors of 2990?

1, 2, 5, 10, 13, 23, 26, 46, 65, 115, 130, 230, 299, 598, 1495, 2990

There is a total of 16 positive divisors.
The sum of these divisors is 6048.
The arithmetic mean is 378.

8 even divisors

2, 10, 26, 46, 130, 230, 598, 2990

8 odd divisors

1, 5, 13, 23, 65, 115, 299, 1495

How to compute the divisors of 2990?

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 2990 by each of the numbers from 1 to 2990 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)

2990 / 1 = 2990 (the remainder is 0, so 1 is a divisor of 2990)
2990 / 2 = 1495 (the remainder is 0, so 2 is a divisor of 2990)
2990 / 3 = 996.66666666667 (the remainder is 2, so 3 is not a divisor of 2990)
...
2990 / 2989 = 1.0003345600535 (the remainder is 1, so 2989 is not a divisor of 2990)
2990 / 2990 = 1 (the remainder is 0, so 2990 is a divisor of 2990)

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 2990 (i.e. 54.680892457969). 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 $\frac{N}{d} = D$ )

2990 / 1 = 2990 (the remainder is 0, so 1 and 2990 are divisors of 2990)
2990 / 2 = 1495 (the remainder is 0, so 2 and 1495 are divisors of 2990)
2990 / 3 = 996.66666666667 (the remainder is 2, so 3 is not a divisor of 2990)
...
2990 / 53 = 56.415094339623 (the remainder is 22, so 53 is not a divisor of 2990)
2990 / 54 = 55.37037037037 (the remainder is 20, so 54 is not a divisor of 2990)