What are the divisors of 2934?

1, 2, 3, 6, 9, 18, 163, 326, 489, 978, 1467, 2934

There is a total of 12 positive divisors.
The sum of these divisors is 6396.
The arithmetic mean is 533.

6 even divisors

2, 6, 18, 326, 978, 2934

6 odd divisors

1, 3, 9, 163, 489, 1467

How to compute the divisors of 2934?

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 2934 by each of the numbers from 1 to 2934 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)

2934 / 1 = 2934 (the remainder is 0, so 1 is a divisor of 2934)
2934 / 2 = 1467 (the remainder is 0, so 2 is a divisor of 2934)
2934 / 3 = 978 (the remainder is 0, so 3 is a divisor of 2934)
...
2934 / 2933 = 1.000340947835 (the remainder is 1, so 2933 is not a divisor of 2934)
2934 / 2934 = 1 (the remainder is 0, so 2934 is a divisor of 2934)

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 2934 (i.e. 54.166410255803). 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 $\frac{N}{d} = D$ )

2934 / 1 = 2934 (the remainder is 0, so 1 and 2934 are divisors of 2934)
2934 / 2 = 1467 (the remainder is 0, so 2 and 1467 are divisors of 2934)
2934 / 3 = 978 (the remainder is 0, so 3 and 978 are divisors of 2934)
...
2934 / 53 = 55.358490566038 (the remainder is 19, so 53 is not a divisor of 2934)
2934 / 54 = 54.333333333333 (the remainder is 18, so 54 is not a divisor of 2934)