What are the numbers divisible by 900?

900, 1800, 2700, 3600, 4500, 5400, 6300, 7200, 8100, 9000, 9900, 10800, 11700, 12600, 13500, 14400, 15300, 16200, 17100, 18000, 18900, 19800, 20700, 21600, 22500, 23400, 24300, 25200, 26100, 27000, 27900, 28800, 29700, 30600, 31500, 32400, 33300, 34200, 35100, 36000, 36900, 37800, 38700, 39600, 40500, 41400, 42300, 43200, 44100, 45000, 45900, 46800, 47700, 48600, 49500, 50400, 51300, 52200, 53100, 54000, 54900, 55800, 56700, 57600, 58500, 59400, 60300, 61200, 62100, 63000, 63900, 64800, 65700, 66600, 67500, 68400, 69300, 70200, 71100, 72000, 72900, 73800, 74700, 75600, 76500, 77400, 78300, 79200, 80100, 81000, 81900, 82800, 83700, 84600, 85500, 86400, 87300, 88200, 89100, 90000, 90900, 91800, 92700, 93600, 94500, 95400, 96300, 97200, 98100, 99000, 99900

There is a total of 111 numbers (up to 100000) that are divisible by 900.
The sum of these numbers is 5594400.
The arithmetic mean of these numbers is 50400.

How to find the numbers divisible by 900?

Finding all the numbers that can be divided by 900 is essentially the same as searching for the multiples of 900: if a number N is a multiple of 900, then 900 is a divisor of N.

Indeed, if we assume that N is a multiple of 900, this means there exists an integer k such that:

$k \times 900 = N$

Conversely, the result of N divided by 900 is this same integer k (without any remainder):

$k = \frac{N}{900}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 900 (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 900 less than 100000):

1 × 900 = 900
2 × 900 = 1800
3 × 900 = 2700
...
110 × 900 = 99000
111 × 900 = 99900