What are the divisors of 2034?

1, 2, 3, 6, 9, 18, 113, 226, 339, 678, 1017, 2034

There is a total of 12 positive divisors.
The sum of these divisors is 4446.
The arithmetic mean is 370.5.

6 even divisors

2, 6, 18, 226, 678, 2034

6 odd divisors

1, 3, 9, 113, 339, 1017

How to compute the divisors of 2034?

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 2034 by each of the numbers from 1 to 2034 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)

2034 / 1 = 2034 (the remainder is 0, so 1 is a divisor of 2034)
2034 / 2 = 1017 (the remainder is 0, so 2 is a divisor of 2034)
2034 / 3 = 678 (the remainder is 0, so 3 is a divisor of 2034)
...
2034 / 2033 = 1.0004918839154 (the remainder is 1, so 2033 is not a divisor of 2034)
2034 / 2034 = 1 (the remainder is 0, so 2034 is a divisor of 2034)

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 2034 (i.e. 45.099889135119). 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 $\frac{N}{d} = D$ )

2034 / 1 = 2034 (the remainder is 0, so 1 and 2034 are divisors of 2034)
2034 / 2 = 1017 (the remainder is 0, so 2 and 1017 are divisors of 2034)
2034 / 3 = 678 (the remainder is 0, so 3 and 678 are divisors of 2034)
...
2034 / 44 = 46.227272727273 (the remainder is 10, so 44 is not a divisor of 2034)
2034 / 45 = 45.2 (the remainder is 9, so 45 is not a divisor of 2034)