What are the numbers divisible by 947?

947, 1894, 2841, 3788, 4735, 5682, 6629, 7576, 8523, 9470, 10417, 11364, 12311, 13258, 14205, 15152, 16099, 17046, 17993, 18940, 19887, 20834, 21781, 22728, 23675, 24622, 25569, 26516, 27463, 28410, 29357, 30304, 31251, 32198, 33145, 34092, 35039, 35986, 36933, 37880, 38827, 39774, 40721, 41668, 42615, 43562, 44509, 45456, 46403, 47350, 48297, 49244, 50191, 51138, 52085, 53032, 53979, 54926, 55873, 56820, 57767, 58714, 59661, 60608, 61555, 62502, 63449, 64396, 65343, 66290, 67237, 68184, 69131, 70078, 71025, 71972, 72919, 73866, 74813, 75760, 76707, 77654, 78601, 79548, 80495, 81442, 82389, 83336, 84283, 85230, 86177, 87124, 88071, 89018, 89965, 90912, 91859, 92806, 93753, 94700, 95647, 96594, 97541, 98488, 99435

There is a total of 105 numbers (up to 100000) that are divisible by 947.
The sum of these numbers is 5270055.
The arithmetic mean of these numbers is 50191.

How to find the numbers divisible by 947?

Finding all the numbers that can be divided by 947 is essentially the same as searching for the multiples of 947: if a number N is a multiple of 947, then 947 is a divisor of N.

Indeed, if we assume that N is a multiple of 947, this means there exists an integer k such that:

$k \times 947 = N$

Conversely, the result of N divided by 947 is this same integer k (without any remainder):

$k = \frac{N}{947}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 947 (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 947 less than 100000):

1 × 947 = 947
2 × 947 = 1894
3 × 947 = 2841
...
104 × 947 = 98488
105 × 947 = 99435