What are the divisors of 2910?

1, 2, 3, 5, 6, 10, 15, 30, 97, 194, 291, 485, 582, 970, 1455, 2910

There is a total of 16 positive divisors.
The sum of these divisors is 7056.
The arithmetic mean is 441.

8 even divisors

2, 6, 10, 30, 194, 582, 970, 2910

8 odd divisors

1, 3, 5, 15, 97, 291, 485, 1455

How to compute the divisors of 2910?

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 2910 by each of the numbers from 1 to 2910 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)

2910 / 1 = 2910 (the remainder is 0, so 1 is a divisor of 2910)
2910 / 2 = 1455 (the remainder is 0, so 2 is a divisor of 2910)
2910 / 3 = 970 (the remainder is 0, so 3 is a divisor of 2910)
...
2910 / 2909 = 1.0003437607425 (the remainder is 1, so 2909 is not a divisor of 2910)
2910 / 2910 = 1 (the remainder is 0, so 2910 is a divisor of 2910)

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 2910 (i.e. 53.944415837045). 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 $\frac{N}{d} = D$ )

2910 / 1 = 2910 (the remainder is 0, so 1 and 2910 are divisors of 2910)
2910 / 2 = 1455 (the remainder is 0, so 2 and 1455 are divisors of 2910)
2910 / 3 = 970 (the remainder is 0, so 3 and 970 are divisors of 2910)
...
2910 / 52 = 55.961538461538 (the remainder is 50, so 52 is not a divisor of 2910)
2910 / 53 = 54.905660377358 (the remainder is 48, so 53 is not a divisor of 2910)