What are the divisors of 3003?

1, 3, 7, 11, 13, 21, 33, 39, 77, 91, 143, 231, 273, 429, 1001, 3003

There is a total of 16 positive divisors.
The sum of these divisors is 5376.
The arithmetic mean is 336.

16 odd divisors

1, 3, 7, 11, 13, 21, 33, 39, 77, 91, 143, 231, 273, 429, 1001, 3003

How to compute the divisors of 3003?

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 3003 by each of the numbers from 1 to 3003 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)

3003 / 1 = 3003 (the remainder is 0, so 1 is a divisor of 3003)
3003 / 2 = 1501.5 (the remainder is 1, so 2 is not a divisor of 3003)
3003 / 3 = 1001 (the remainder is 0, so 3 is a divisor of 3003)
...
3003 / 3002 = 1.0003331112592 (the remainder is 1, so 3002 is not a divisor of 3003)
3003 / 3003 = 1 (the remainder is 0, so 3003 is a divisor of 3003)

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 3003 (i.e. 54.799635035281). 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 $\frac{N}{d} = D$ )

3003 / 1 = 3003 (the remainder is 0, so 1 and 3003 are divisors of 3003)
3003 / 2 = 1501.5 (the remainder is 1, so 2 is not a divisor of 3003)
3003 / 3 = 1001 (the remainder is 0, so 3 and 1001 are divisors of 3003)
...
3003 / 53 = 56.660377358491 (the remainder is 35, so 53 is not a divisor of 3003)
3003 / 54 = 55.611111111111 (the remainder is 33, so 54 is not a divisor of 3003)