What are the numbers divisible by 938?

938, 1876, 2814, 3752, 4690, 5628, 6566, 7504, 8442, 9380, 10318, 11256, 12194, 13132, 14070, 15008, 15946, 16884, 17822, 18760, 19698, 20636, 21574, 22512, 23450, 24388, 25326, 26264, 27202, 28140, 29078, 30016, 30954, 31892, 32830, 33768, 34706, 35644, 36582, 37520, 38458, 39396, 40334, 41272, 42210, 43148, 44086, 45024, 45962, 46900, 47838, 48776, 49714, 50652, 51590, 52528, 53466, 54404, 55342, 56280, 57218, 58156, 59094, 60032, 60970, 61908, 62846, 63784, 64722, 65660, 66598, 67536, 68474, 69412, 70350, 71288, 72226, 73164, 74102, 75040, 75978, 76916, 77854, 78792, 79730, 80668, 81606, 82544, 83482, 84420, 85358, 86296, 87234, 88172, 89110, 90048, 90986, 91924, 92862, 93800, 94738, 95676, 96614, 97552, 98490, 99428

There is a total of 106 numbers (up to 100000) that are divisible by 938.
The sum of these numbers is 5319398.
The arithmetic mean of these numbers is 50183.

How to find the numbers divisible by 938?

Finding all the numbers that can be divided by 938 is essentially the same as searching for the multiples of 938: if a number N is a multiple of 938, then 938 is a divisor of N.

Indeed, if we assume that N is a multiple of 938, this means there exists an integer k such that:

$k \times 938 = N$

Conversely, the result of N divided by 938 is this same integer k (without any remainder):

$k = \frac{N}{938}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 938 (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 938 less than 100000):

1 × 938 = 938
2 × 938 = 1876
3 × 938 = 2814
...
105 × 938 = 98490
106 × 938 = 99428