What are the divisors of 2490?

1, 2, 3, 5, 6, 10, 15, 30, 83, 166, 249, 415, 498, 830, 1245, 2490

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

8 even divisors

2, 6, 10, 30, 166, 498, 830, 2490

8 odd divisors

1, 3, 5, 15, 83, 249, 415, 1245

How to compute the divisors of 2490?

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 2490 by each of the numbers from 1 to 2490 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)

2490 / 1 = 2490 (the remainder is 0, so 1 is a divisor of 2490)
2490 / 2 = 1245 (the remainder is 0, so 2 is a divisor of 2490)
2490 / 3 = 830 (the remainder is 0, so 3 is a divisor of 2490)
...
2490 / 2489 = 1.0004017677782 (the remainder is 1, so 2489 is not a divisor of 2490)
2490 / 2490 = 1 (the remainder is 0, so 2490 is a divisor of 2490)

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 2490 (i.e. 49.899899799499). 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 $\frac{N}{d} = D$ )

2490 / 1 = 2490 (the remainder is 0, so 1 and 2490 are divisors of 2490)
2490 / 2 = 1245 (the remainder is 0, so 2 and 1245 are divisors of 2490)
2490 / 3 = 830 (the remainder is 0, so 3 and 830 are divisors of 2490)
...
2490 / 48 = 51.875 (the remainder is 42, so 48 is not a divisor of 2490)
2490 / 49 = 50.816326530612 (the remainder is 40, so 49 is not a divisor of 2490)