What are the divisors of 8052?

1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 61, 66, 122, 132, 183, 244, 366, 671, 732, 1342, 2013, 2684, 4026, 8052

There is a total of 24 positive divisors.
The sum of these divisors is 20832.
The arithmetic mean is 868.

16 even divisors

2, 4, 6, 12, 22, 44, 66, 122, 132, 244, 366, 732, 1342, 2684, 4026, 8052

8 odd divisors

1, 3, 11, 33, 61, 183, 671, 2013

How to compute the divisors of 8052?

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 8052 by each of the numbers from 1 to 8052 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)

8052 / 1 = 8052 (the remainder is 0, so 1 is a divisor of 8052)
8052 / 2 = 4026 (the remainder is 0, so 2 is a divisor of 8052)
8052 / 3 = 2684 (the remainder is 0, so 3 is a divisor of 8052)
...
8052 / 8051 = 1.0001242081729 (the remainder is 1, so 8051 is not a divisor of 8052)
8052 / 8052 = 1 (the remainder is 0, so 8052 is a divisor of 8052)

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 8052 (i.e. 89.732937096698). 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 $\frac{N}{d} = D$ )

8052 / 1 = 8052 (the remainder is 0, so 1 and 8052 are divisors of 8052)
8052 / 2 = 4026 (the remainder is 0, so 2 and 4026 are divisors of 8052)
8052 / 3 = 2684 (the remainder is 0, so 3 and 2684 are divisors of 8052)
...
8052 / 88 = 91.5 (the remainder is 44, so 88 is not a divisor of 8052)
8052 / 89 = 90.47191011236 (the remainder is 42, so 89 is not a divisor of 8052)