What are the numbers divisible by 1040?

1040, 2080, 3120, 4160, 5200, 6240, 7280, 8320, 9360, 10400, 11440, 12480, 13520, 14560, 15600, 16640, 17680, 18720, 19760, 20800, 21840, 22880, 23920, 24960, 26000, 27040, 28080, 29120, 30160, 31200, 32240, 33280, 34320, 35360, 36400, 37440, 38480, 39520, 40560, 41600, 42640, 43680, 44720, 45760, 46800, 47840, 48880, 49920, 50960, 52000, 53040, 54080, 55120, 56160, 57200, 58240, 59280, 60320, 61360, 62400, 63440, 64480, 65520, 66560, 67600, 68640, 69680, 70720, 71760, 72800, 73840, 74880, 75920, 76960, 78000, 79040, 80080, 81120, 82160, 83200, 84240, 85280, 86320, 87360, 88400, 89440, 90480, 91520, 92560, 93600, 94640, 95680, 96720, 97760, 98800, 99840

There is a total of 96 numbers (up to 100000) that are divisible by 1040.
The sum of these numbers is 4842240.
The arithmetic mean of these numbers is 50440.

How to find the numbers divisible by 1040?

Finding all the numbers that can be divided by 1040 is essentially the same as searching for the multiples of 1040: if a number N is a multiple of 1040, then 1040 is a divisor of N.

Indeed, if we assume that N is a multiple of 1040, this means there exists an integer k such that:

$k \times 1040 = N$

Conversely, the result of N divided by 1040 is this same integer k (without any remainder):

$k = \frac{N}{1040}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 1040 (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 1040 less than 100000):

1 × 1040 = 1040
2 × 1040 = 2080
3 × 1040 = 3120
...
95 × 1040 = 98800
96 × 1040 = 99840