What are the numbers divisible by 1003?

1003, 2006, 3009, 4012, 5015, 6018, 7021, 8024, 9027, 10030, 11033, 12036, 13039, 14042, 15045, 16048, 17051, 18054, 19057, 20060, 21063, 22066, 23069, 24072, 25075, 26078, 27081, 28084, 29087, 30090, 31093, 32096, 33099, 34102, 35105, 36108, 37111, 38114, 39117, 40120, 41123, 42126, 43129, 44132, 45135, 46138, 47141, 48144, 49147, 50150, 51153, 52156, 53159, 54162, 55165, 56168, 57171, 58174, 59177, 60180, 61183, 62186, 63189, 64192, 65195, 66198, 67201, 68204, 69207, 70210, 71213, 72216, 73219, 74222, 75225, 76228, 77231, 78234, 79237, 80240, 81243, 82246, 83249, 84252, 85255, 86258, 87261, 88264, 89267, 90270, 91273, 92276, 93279, 94282, 95285, 96288, 97291, 98294, 99297

There is a total of 99 numbers (up to 100000) that are divisible by 1003.
The sum of these numbers is 4964850.
The arithmetic mean of these numbers is 50150.

How to find the numbers divisible by 1003?

Finding all the numbers that can be divided by 1003 is essentially the same as searching for the multiples of 1003: if a number N is a multiple of 1003, then 1003 is a divisor of N.

Indeed, if we assume that N is a multiple of 1003, this means there exists an integer k such that:

$k \times 1003 = N$

Conversely, the result of N divided by 1003 is this same integer k (without any remainder):

$k = \frac{N}{1003}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 1003 (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 1003 less than 100000):

1 × 1003 = 1003
2 × 1003 = 2006
3 × 1003 = 3009
...
98 × 1003 = 98294
99 × 1003 = 99297