What are the numbers divisible by 918?

918, 1836, 2754, 3672, 4590, 5508, 6426, 7344, 8262, 9180, 10098, 11016, 11934, 12852, 13770, 14688, 15606, 16524, 17442, 18360, 19278, 20196, 21114, 22032, 22950, 23868, 24786, 25704, 26622, 27540, 28458, 29376, 30294, 31212, 32130, 33048, 33966, 34884, 35802, 36720, 37638, 38556, 39474, 40392, 41310, 42228, 43146, 44064, 44982, 45900, 46818, 47736, 48654, 49572, 50490, 51408, 52326, 53244, 54162, 55080, 55998, 56916, 57834, 58752, 59670, 60588, 61506, 62424, 63342, 64260, 65178, 66096, 67014, 67932, 68850, 69768, 70686, 71604, 72522, 73440, 74358, 75276, 76194, 77112, 78030, 78948, 79866, 80784, 81702, 82620, 83538, 84456, 85374, 86292, 87210, 88128, 89046, 89964, 90882, 91800, 92718, 93636, 94554, 95472, 96390, 97308, 98226, 99144

There is a total of 108 numbers (up to 100000) that are divisible by 918.
The sum of these numbers is 5403348.
The arithmetic mean of these numbers is 50031.

How to find the numbers divisible by 918?

Finding all the numbers that can be divided by 918 is essentially the same as searching for the multiples of 918: if a number N is a multiple of 918, then 918 is a divisor of N.

Indeed, if we assume that N is a multiple of 918, this means there exists an integer k such that:

$k \times 918 = N$

Conversely, the result of N divided by 918 is this same integer k (without any remainder):

$k = \frac{N}{918}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 918 (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 918 less than 100000):

1 × 918 = 918
2 × 918 = 1836
3 × 918 = 2754
...
107 × 918 = 98226
108 × 918 = 99144