What are the divisors of 2970?

1, 2, 3, 5, 6, 9, 10, 11, 15, 18, 22, 27, 30, 33, 45, 54, 55, 66, 90, 99, 110, 135, 165, 198, 270, 297, 330, 495, 594, 990, 1485, 2970

There is a total of 32 positive divisors.
The sum of these divisors is 8640.
The arithmetic mean is 270.

16 even divisors

2, 6, 10, 18, 22, 30, 54, 66, 90, 110, 198, 270, 330, 594, 990, 2970

16 odd divisors

1, 3, 5, 9, 11, 15, 27, 33, 45, 55, 99, 135, 165, 297, 495, 1485

How to compute the divisors of 2970?

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 2970 by each of the numbers from 1 to 2970 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)

2970 / 1 = 2970 (the remainder is 0, so 1 is a divisor of 2970)
2970 / 2 = 1485 (the remainder is 0, so 2 is a divisor of 2970)
2970 / 3 = 990 (the remainder is 0, so 3 is a divisor of 2970)
...
2970 / 2969 = 1.000336813742 (the remainder is 1, so 2969 is not a divisor of 2970)
2970 / 2970 = 1 (the remainder is 0, so 2970 is a divisor of 2970)

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 2970 (i.e. 54.497706373755). 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 $\frac{N}{d} = D$ )

2970 / 1 = 2970 (the remainder is 0, so 1 and 2970 are divisors of 2970)
2970 / 2 = 1485 (the remainder is 0, so 2 and 1485 are divisors of 2970)
2970 / 3 = 990 (the remainder is 0, so 3 and 990 are divisors of 2970)
...
2970 / 53 = 56.037735849057 (the remainder is 2, so 53 is not a divisor of 2970)
2970 / 54 = 55 (the remainder is 0, so 54 and 55 are divisors of 2970)