What are the divisors of 2907?

1, 3, 9, 17, 19, 51, 57, 153, 171, 323, 969, 2907

There is a total of 12 positive divisors.
The sum of these divisors is 4680.
The arithmetic mean is 390.

12 odd divisors

1, 3, 9, 17, 19, 51, 57, 153, 171, 323, 969, 2907

How to compute the divisors of 2907?

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 2907 by each of the numbers from 1 to 2907 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)

2907 / 1 = 2907 (the remainder is 0, so 1 is a divisor of 2907)
2907 / 2 = 1453.5 (the remainder is 1, so 2 is not a divisor of 2907)
2907 / 3 = 969 (the remainder is 0, so 3 is a divisor of 2907)
...
2907 / 2906 = 1.0003441156228 (the remainder is 1, so 2906 is not a divisor of 2907)
2907 / 2907 = 1 (the remainder is 0, so 2907 is a divisor of 2907)

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 2907 (i.e. 53.916602266834). 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 $\frac{N}{d} = D$ )

2907 / 1 = 2907 (the remainder is 0, so 1 and 2907 are divisors of 2907)
2907 / 2 = 1453.5 (the remainder is 1, so 2 is not a divisor of 2907)
2907 / 3 = 969 (the remainder is 0, so 3 and 969 are divisors of 2907)
...
2907 / 52 = 55.903846153846 (the remainder is 47, so 52 is not a divisor of 2907)
2907 / 53 = 54.849056603774 (the remainder is 45, so 53 is not a divisor of 2907)