What are the numbers divisible by 941?

941, 1882, 2823, 3764, 4705, 5646, 6587, 7528, 8469, 9410, 10351, 11292, 12233, 13174, 14115, 15056, 15997, 16938, 17879, 18820, 19761, 20702, 21643, 22584, 23525, 24466, 25407, 26348, 27289, 28230, 29171, 30112, 31053, 31994, 32935, 33876, 34817, 35758, 36699, 37640, 38581, 39522, 40463, 41404, 42345, 43286, 44227, 45168, 46109, 47050, 47991, 48932, 49873, 50814, 51755, 52696, 53637, 54578, 55519, 56460, 57401, 58342, 59283, 60224, 61165, 62106, 63047, 63988, 64929, 65870, 66811, 67752, 68693, 69634, 70575, 71516, 72457, 73398, 74339, 75280, 76221, 77162, 78103, 79044, 79985, 80926, 81867, 82808, 83749, 84690, 85631, 86572, 87513, 88454, 89395, 90336, 91277, 92218, 93159, 94100, 95041, 95982, 96923, 97864, 98805, 99746

There is a total of 106 numbers (up to 100000) that are divisible by 941.
The sum of these numbers is 5336411.
The arithmetic mean of these numbers is 50343.5.

How to find the numbers divisible by 941?

Finding all the numbers that can be divided by 941 is essentially the same as searching for the multiples of 941: if a number N is a multiple of 941, then 941 is a divisor of N.

Indeed, if we assume that N is a multiple of 941, this means there exists an integer k such that:

$k \times 941 = N$

Conversely, the result of N divided by 941 is this same integer k (without any remainder):

$k = \frac{N}{941}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 941 (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 941 less than 100000):

1 × 941 = 941
2 × 941 = 1882
3 × 941 = 2823
...
105 × 941 = 98805
106 × 941 = 99746