What are the divisors of 2208?

1, 2, 3, 4, 6, 8, 12, 16, 23, 24, 32, 46, 48, 69, 92, 96, 138, 184, 276, 368, 552, 736, 1104, 2208

There is a total of 24 positive divisors.
The sum of these divisors is 6048.
The arithmetic mean is 252.

20 even divisors

2, 4, 6, 8, 12, 16, 24, 32, 46, 48, 92, 96, 138, 184, 276, 368, 552, 736, 1104, 2208

4 odd divisors

1, 3, 23, 69

How to compute the divisors of 2208?

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 2208 by each of the numbers from 1 to 2208 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)

2208 / 1 = 2208 (the remainder is 0, so 1 is a divisor of 2208)
2208 / 2 = 1104 (the remainder is 0, so 2 is a divisor of 2208)
2208 / 3 = 736 (the remainder is 0, so 3 is a divisor of 2208)
...
2208 / 2207 = 1.0004531037608 (the remainder is 1, so 2207 is not a divisor of 2208)
2208 / 2208 = 1 (the remainder is 0, so 2208 is a divisor of 2208)

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 2208 (i.e. 46.989360497883). 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 $\frac{N}{d} = D$ )

2208 / 1 = 2208 (the remainder is 0, so 1 and 2208 are divisors of 2208)
2208 / 2 = 1104 (the remainder is 0, so 2 and 1104 are divisors of 2208)
2208 / 3 = 736 (the remainder is 0, so 3 and 736 are divisors of 2208)
...
2208 / 45 = 49.066666666667 (the remainder is 3, so 45 is not a divisor of 2208)
2208 / 46 = 48 (the remainder is 0, so 46 and 48 are divisors of 2208)