What are the numbers divisible by 2005?

2005, 4010, 6015, 8020, 10025, 12030, 14035, 16040, 18045, 20050, 22055, 24060, 26065, 28070, 30075, 32080, 34085, 36090, 38095, 40100, 42105, 44110, 46115, 48120, 50125, 52130, 54135, 56140, 58145, 60150, 62155, 64160, 66165, 68170, 70175, 72180, 74185, 76190, 78195, 80200, 82205, 84210, 86215, 88220, 90225, 92230, 94235, 96240, 98245

There is a total of 49 numbers (up to 100000) that are divisible by 2005.
The sum of these numbers is 2456125.
The arithmetic mean of these numbers is 50125.

How to find the numbers divisible by 2005?

Finding all the numbers that can be divided by 2005 is essentially the same as searching for the multiples of 2005: if a number N is a multiple of 2005, then 2005 is a divisor of N.

Indeed, if we assume that N is a multiple of 2005, this means there exists an integer k such that:

$k \times 2005 = N$

Conversely, the result of N divided by 2005 is this same integer k (without any remainder):

$k = \frac{N}{2005}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 2005 (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 2005 less than 100000):

1 × 2005 = 2005
2 × 2005 = 4010
3 × 2005 = 6015
...
48 × 2005 = 96240
49 × 2005 = 98245