What are the numbers divisible by 1002?

1002, 2004, 3006, 4008, 5010, 6012, 7014, 8016, 9018, 10020, 11022, 12024, 13026, 14028, 15030, 16032, 17034, 18036, 19038, 20040, 21042, 22044, 23046, 24048, 25050, 26052, 27054, 28056, 29058, 30060, 31062, 32064, 33066, 34068, 35070, 36072, 37074, 38076, 39078, 40080, 41082, 42084, 43086, 44088, 45090, 46092, 47094, 48096, 49098, 50100, 51102, 52104, 53106, 54108, 55110, 56112, 57114, 58116, 59118, 60120, 61122, 62124, 63126, 64128, 65130, 66132, 67134, 68136, 69138, 70140, 71142, 72144, 73146, 74148, 75150, 76152, 77154, 78156, 79158, 80160, 81162, 82164, 83166, 84168, 85170, 86172, 87174, 88176, 89178, 90180, 91182, 92184, 93186, 94188, 95190, 96192, 97194, 98196, 99198

There is a total of 99 numbers (up to 100000) that are divisible by 1002.
The sum of these numbers is 4959900.
The arithmetic mean of these numbers is 50100.

How to find the numbers divisible by 1002?

Finding all the numbers that can be divided by 1002 is essentially the same as searching for the multiples of 1002: if a number N is a multiple of 1002, then 1002 is a divisor of N.

Indeed, if we assume that N is a multiple of 1002, this means there exists an integer k such that:

$k \times 1002 = N$

Conversely, the result of N divided by 1002 is this same integer k (without any remainder):

$k = \frac{N}{1002}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 1002 (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 1002 less than 100000):

1 × 1002 = 1002
2 × 1002 = 2004
3 × 1002 = 3006
...
98 × 1002 = 98196
99 × 1002 = 99198