What are the divisors of 2223?

1, 3, 9, 13, 19, 39, 57, 117, 171, 247, 741, 2223

There is a total of 12 positive divisors.
The sum of these divisors is 3640.
The arithmetic mean is 303.33333333333.

12 odd divisors

1, 3, 9, 13, 19, 39, 57, 117, 171, 247, 741, 2223

How to compute the divisors of 2223?

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 2223 by each of the numbers from 1 to 2223 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)

2223 / 1 = 2223 (the remainder is 0, so 1 is a divisor of 2223)
2223 / 2 = 1111.5 (the remainder is 1, so 2 is not a divisor of 2223)
2223 / 3 = 741 (the remainder is 0, so 3 is a divisor of 2223)
...
2223 / 2222 = 1.0004500450045 (the remainder is 1, so 2222 is not a divisor of 2223)
2223 / 2223 = 1 (the remainder is 0, so 2223 is a divisor of 2223)

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 2223 (i.e. 47.148700936505). 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 $\frac{N}{d} = D$ )

2223 / 1 = 2223 (the remainder is 0, so 1 and 2223 are divisors of 2223)
2223 / 2 = 1111.5 (the remainder is 1, so 2 is not a divisor of 2223)
2223 / 3 = 741 (the remainder is 0, so 3 and 741 are divisors of 2223)
...
2223 / 46 = 48.326086956522 (the remainder is 15, so 46 is not a divisor of 2223)
2223 / 47 = 47.297872340426 (the remainder is 14, so 47 is not a divisor of 2223)