What are the numbers divisible by 845?

845, 1690, 2535, 3380, 4225, 5070, 5915, 6760, 7605, 8450, 9295, 10140, 10985, 11830, 12675, 13520, 14365, 15210, 16055, 16900, 17745, 18590, 19435, 20280, 21125, 21970, 22815, 23660, 24505, 25350, 26195, 27040, 27885, 28730, 29575, 30420, 31265, 32110, 32955, 33800, 34645, 35490, 36335, 37180, 38025, 38870, 39715, 40560, 41405, 42250, 43095, 43940, 44785, 45630, 46475, 47320, 48165, 49010, 49855, 50700, 51545, 52390, 53235, 54080, 54925, 55770, 56615, 57460, 58305, 59150, 59995, 60840, 61685, 62530, 63375, 64220, 65065, 65910, 66755, 67600, 68445, 69290, 70135, 70980, 71825, 72670, 73515, 74360, 75205, 76050, 76895, 77740, 78585, 79430, 80275, 81120, 81965, 82810, 83655, 84500, 85345, 86190, 87035, 87880, 88725, 89570, 90415, 91260, 92105, 92950, 93795, 94640, 95485, 96330, 97175, 98020, 98865, 99710

There is a total of 118 numbers (up to 100000) that are divisible by 845.
The sum of these numbers is 5932745.
The arithmetic mean of these numbers is 50277.5.

How to find the numbers divisible by 845?

Finding all the numbers that can be divided by 845 is essentially the same as searching for the multiples of 845: if a number N is a multiple of 845, then 845 is a divisor of N.

Indeed, if we assume that N is a multiple of 845, this means there exists an integer k such that:

$k \times 845 = N$

Conversely, the result of N divided by 845 is this same integer k (without any remainder):

$k = \frac{N}{845}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 845 (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 845 less than 100000):

1 × 845 = 845
2 × 845 = 1690
3 × 845 = 2535
...
117 × 845 = 98865
118 × 845 = 99710