What are the numbers divisible by 990?

990, 1980, 2970, 3960, 4950, 5940, 6930, 7920, 8910, 9900, 10890, 11880, 12870, 13860, 14850, 15840, 16830, 17820, 18810, 19800, 20790, 21780, 22770, 23760, 24750, 25740, 26730, 27720, 28710, 29700, 30690, 31680, 32670, 33660, 34650, 35640, 36630, 37620, 38610, 39600, 40590, 41580, 42570, 43560, 44550, 45540, 46530, 47520, 48510, 49500, 50490, 51480, 52470, 53460, 54450, 55440, 56430, 57420, 58410, 59400, 60390, 61380, 62370, 63360, 64350, 65340, 66330, 67320, 68310, 69300, 70290, 71280, 72270, 73260, 74250, 75240, 76230, 77220, 78210, 79200, 80190, 81180, 82170, 83160, 84150, 85140, 86130, 87120, 88110, 89100, 90090, 91080, 92070, 93060, 94050, 95040, 96030, 97020, 98010, 99000, 99990

There is a total of 101 numbers (up to 100000) that are divisible by 990.
The sum of these numbers is 5099490.
The arithmetic mean of these numbers is 50490.

How to find the numbers divisible by 990?

Finding all the numbers that can be divided by 990 is essentially the same as searching for the multiples of 990: if a number N is a multiple of 990, then 990 is a divisor of N.

Indeed, if we assume that N is a multiple of 990, this means there exists an integer k such that:

$k \times 990 = N$

Conversely, the result of N divided by 990 is this same integer k (without any remainder):

$k = \frac{N}{990}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 990 (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 990 less than 100000):

1 × 990 = 990
2 × 990 = 1980
3 × 990 = 2970
...
100 × 990 = 99000
101 × 990 = 99990