What are the divisors of 9009?

1, 3, 7, 9, 11, 13, 21, 33, 39, 63, 77, 91, 99, 117, 143, 231, 273, 429, 693, 819, 1001, 1287, 3003, 9009

There is a total of 24 positive divisors.
The sum of these divisors is 17472.
The arithmetic mean is 728.

24 odd divisors

1, 3, 7, 9, 11, 13, 21, 33, 39, 63, 77, 91, 99, 117, 143, 231, 273, 429, 693, 819, 1001, 1287, 3003, 9009

How to compute the divisors of 9009?

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 9009 by each of the numbers from 1 to 9009 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)

9009 / 1 = 9009 (the remainder is 0, so 1 is a divisor of 9009)
9009 / 2 = 4504.5 (the remainder is 1, so 2 is not a divisor of 9009)
9009 / 3 = 3003 (the remainder is 0, so 3 is a divisor of 9009)
...
9009 / 9008 = 1.0001110124334 (the remainder is 1, so 9008 is not a divisor of 9009)
9009 / 9009 = 1 (the remainder is 0, so 9009 is a divisor of 9009)

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 9009 (i.e. 94.915752117338). 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 $\frac{N}{d} = D$ )

9009 / 1 = 9009 (the remainder is 0, so 1 and 9009 are divisors of 9009)
9009 / 2 = 4504.5 (the remainder is 1, so 2 is not a divisor of 9009)
9009 / 3 = 3003 (the remainder is 0, so 3 and 3003 are divisors of 9009)
...
9009 / 93 = 96.870967741935 (the remainder is 81, so 93 is not a divisor of 9009)
9009 / 94 = 95.840425531915 (the remainder is 79, so 94 is not a divisor of 9009)