What are the numbers divisible by 945?

945, 1890, 2835, 3780, 4725, 5670, 6615, 7560, 8505, 9450, 10395, 11340, 12285, 13230, 14175, 15120, 16065, 17010, 17955, 18900, 19845, 20790, 21735, 22680, 23625, 24570, 25515, 26460, 27405, 28350, 29295, 30240, 31185, 32130, 33075, 34020, 34965, 35910, 36855, 37800, 38745, 39690, 40635, 41580, 42525, 43470, 44415, 45360, 46305, 47250, 48195, 49140, 50085, 51030, 51975, 52920, 53865, 54810, 55755, 56700, 57645, 58590, 59535, 60480, 61425, 62370, 63315, 64260, 65205, 66150, 67095, 68040, 68985, 69930, 70875, 71820, 72765, 73710, 74655, 75600, 76545, 77490, 78435, 79380, 80325, 81270, 82215, 83160, 84105, 85050, 85995, 86940, 87885, 88830, 89775, 90720, 91665, 92610, 93555, 94500, 95445, 96390, 97335, 98280, 99225

There is a total of 105 numbers (up to 100000) that are divisible by 945.
The sum of these numbers is 5258925.
The arithmetic mean of these numbers is 50085.

How to find the numbers divisible by 945?

Finding all the numbers that can be divided by 945 is essentially the same as searching for the multiples of 945: if a number N is a multiple of 945, then 945 is a divisor of N.

Indeed, if we assume that N is a multiple of 945, this means there exists an integer k such that:

$k \times 945 = N$

Conversely, the result of N divided by 945 is this same integer k (without any remainder):

$k = \frac{N}{945}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 945 (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 945 less than 100000):

1 × 945 = 945
2 × 945 = 1890
3 × 945 = 2835
...
104 × 945 = 98280
105 × 945 = 99225