What are the numbers divisible by 2003?

2003, 4006, 6009, 8012, 10015, 12018, 14021, 16024, 18027, 20030, 22033, 24036, 26039, 28042, 30045, 32048, 34051, 36054, 38057, 40060, 42063, 44066, 46069, 48072, 50075, 52078, 54081, 56084, 58087, 60090, 62093, 64096, 66099, 68102, 70105, 72108, 74111, 76114, 78117, 80120, 82123, 84126, 86129, 88132, 90135, 92138, 94141, 96144, 98147

There is a total of 49 numbers (up to 100000) that are divisible by 2003.
The sum of these numbers is 2453675.
The arithmetic mean of these numbers is 50075.

How to find the numbers divisible by 2003?

Finding all the numbers that can be divided by 2003 is essentially the same as searching for the multiples of 2003: if a number N is a multiple of 2003, then 2003 is a divisor of N.

Indeed, if we assume that N is a multiple of 2003, this means there exists an integer k such that:

$k \times 2003 = N$

Conversely, the result of N divided by 2003 is this same integer k (without any remainder):

$k = \frac{N}{2003}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 2003 (we can keep multiplying it by increasingly larger integers, without ever reaching the end).

However, in this instance, we've chosen to set an arbitrary limit (specifically, the multiples of 2003 less than 100000):

1 × 2003 = 2003
2 × 2003 = 4006
3 × 2003 = 6009
...
48 × 2003 = 96144
49 × 2003 = 98147