What are the numbers divisible by 933?

933, 1866, 2799, 3732, 4665, 5598, 6531, 7464, 8397, 9330, 10263, 11196, 12129, 13062, 13995, 14928, 15861, 16794, 17727, 18660, 19593, 20526, 21459, 22392, 23325, 24258, 25191, 26124, 27057, 27990, 28923, 29856, 30789, 31722, 32655, 33588, 34521, 35454, 36387, 37320, 38253, 39186, 40119, 41052, 41985, 42918, 43851, 44784, 45717, 46650, 47583, 48516, 49449, 50382, 51315, 52248, 53181, 54114, 55047, 55980, 56913, 57846, 58779, 59712, 60645, 61578, 62511, 63444, 64377, 65310, 66243, 67176, 68109, 69042, 69975, 70908, 71841, 72774, 73707, 74640, 75573, 76506, 77439, 78372, 79305, 80238, 81171, 82104, 83037, 83970, 84903, 85836, 86769, 87702, 88635, 89568, 90501, 91434, 92367, 93300, 94233, 95166, 96099, 97032, 97965, 98898, 99831

There is a total of 107 numbers (up to 100000) that are divisible by 933.
The sum of these numbers is 5390874.
The arithmetic mean of these numbers is 50382.

How to find the numbers divisible by 933?

Finding all the numbers that can be divided by 933 is essentially the same as searching for the multiples of 933: if a number N is a multiple of 933, then 933 is a divisor of N.

Indeed, if we assume that N is a multiple of 933, this means there exists an integer k such that:

$k \times 933 = N$

Conversely, the result of N divided by 933 is this same integer k (without any remainder):

$k = \frac{N}{933}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 933 (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 933 less than 100000):

1 × 933 = 933
2 × 933 = 1866
3 × 933 = 2799
...
106 × 933 = 98898
107 × 933 = 99831