What are the numbers divisible by 915?

915, 1830, 2745, 3660, 4575, 5490, 6405, 7320, 8235, 9150, 10065, 10980, 11895, 12810, 13725, 14640, 15555, 16470, 17385, 18300, 19215, 20130, 21045, 21960, 22875, 23790, 24705, 25620, 26535, 27450, 28365, 29280, 30195, 31110, 32025, 32940, 33855, 34770, 35685, 36600, 37515, 38430, 39345, 40260, 41175, 42090, 43005, 43920, 44835, 45750, 46665, 47580, 48495, 49410, 50325, 51240, 52155, 53070, 53985, 54900, 55815, 56730, 57645, 58560, 59475, 60390, 61305, 62220, 63135, 64050, 64965, 65880, 66795, 67710, 68625, 69540, 70455, 71370, 72285, 73200, 74115, 75030, 75945, 76860, 77775, 78690, 79605, 80520, 81435, 82350, 83265, 84180, 85095, 86010, 86925, 87840, 88755, 89670, 90585, 91500, 92415, 93330, 94245, 95160, 96075, 96990, 97905, 98820, 99735

There is a total of 109 numbers (up to 100000) that are divisible by 915.
The sum of these numbers is 5485425.
The arithmetic mean of these numbers is 50325.

How to find the numbers divisible by 915?

Finding all the numbers that can be divided by 915 is essentially the same as searching for the multiples of 915: if a number N is a multiple of 915, then 915 is a divisor of N.

Indeed, if we assume that N is a multiple of 915, this means there exists an integer k such that:

$k \times 915 = N$

Conversely, the result of N divided by 915 is this same integer k (without any remainder):

$k = \frac{N}{915}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 915 (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 915 less than 100000):

1 × 915 = 915
2 × 915 = 1830
3 × 915 = 2745
...
108 × 915 = 98820
109 × 915 = 99735