What are the numbers divisible by 936?

936, 1872, 2808, 3744, 4680, 5616, 6552, 7488, 8424, 9360, 10296, 11232, 12168, 13104, 14040, 14976, 15912, 16848, 17784, 18720, 19656, 20592, 21528, 22464, 23400, 24336, 25272, 26208, 27144, 28080, 29016, 29952, 30888, 31824, 32760, 33696, 34632, 35568, 36504, 37440, 38376, 39312, 40248, 41184, 42120, 43056, 43992, 44928, 45864, 46800, 47736, 48672, 49608, 50544, 51480, 52416, 53352, 54288, 55224, 56160, 57096, 58032, 58968, 59904, 60840, 61776, 62712, 63648, 64584, 65520, 66456, 67392, 68328, 69264, 70200, 71136, 72072, 73008, 73944, 74880, 75816, 76752, 77688, 78624, 79560, 80496, 81432, 82368, 83304, 84240, 85176, 86112, 87048, 87984, 88920, 89856, 90792, 91728, 92664, 93600, 94536, 95472, 96408, 97344, 98280, 99216

There is a total of 106 numbers (up to 100000) that are divisible by 936.
The sum of these numbers is 5308056.
The arithmetic mean of these numbers is 50076.

How to find the numbers divisible by 936?

Finding all the numbers that can be divided by 936 is essentially the same as searching for the multiples of 936: if a number N is a multiple of 936, then 936 is a divisor of N.

Indeed, if we assume that N is a multiple of 936, this means there exists an integer k such that:

$k \times 936 = N$

Conversely, the result of N divided by 936 is this same integer k (without any remainder):

$k = \frac{N}{936}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 936 (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 936 less than 100000):

1 × 936 = 936
2 × 936 = 1872
3 × 936 = 2808
...
105 × 936 = 98280
106 × 936 = 99216