What are the divisors of 2652?

1, 2, 3, 4, 6, 12, 13, 17, 26, 34, 39, 51, 52, 68, 78, 102, 156, 204, 221, 442, 663, 884, 1326, 2652

There is a total of 24 positive divisors.
The sum of these divisors is 7056.
The arithmetic mean is 294.

16 even divisors

2, 4, 6, 12, 26, 34, 52, 68, 78, 102, 156, 204, 442, 884, 1326, 2652

8 odd divisors

1, 3, 13, 17, 39, 51, 221, 663

How to compute the divisors of 2652?

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 2652 by each of the numbers from 1 to 2652 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)

2652 / 1 = 2652 (the remainder is 0, so 1 is a divisor of 2652)
2652 / 2 = 1326 (the remainder is 0, so 2 is a divisor of 2652)
2652 / 3 = 884 (the remainder is 0, so 3 is a divisor of 2652)
...
2652 / 2651 = 1.0003772161449 (the remainder is 1, so 2651 is not a divisor of 2652)
2652 / 2652 = 1 (the remainder is 0, so 2652 is a divisor of 2652)

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 2652 (i.e. 51.497572758335). 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 $\frac{N}{d} = D$ )

2652 / 1 = 2652 (the remainder is 0, so 1 and 2652 are divisors of 2652)
2652 / 2 = 1326 (the remainder is 0, so 2 and 1326 are divisors of 2652)
2652 / 3 = 884 (the remainder is 0, so 3 and 884 are divisors of 2652)
...
2652 / 50 = 53.04 (the remainder is 2, so 50 is not a divisor of 2652)
2652 / 51 = 52 (the remainder is 0, so 51 and 52 are divisors of 2652)