What are the numbers divisible by 880?

880, 1760, 2640, 3520, 4400, 5280, 6160, 7040, 7920, 8800, 9680, 10560, 11440, 12320, 13200, 14080, 14960, 15840, 16720, 17600, 18480, 19360, 20240, 21120, 22000, 22880, 23760, 24640, 25520, 26400, 27280, 28160, 29040, 29920, 30800, 31680, 32560, 33440, 34320, 35200, 36080, 36960, 37840, 38720, 39600, 40480, 41360, 42240, 43120, 44000, 44880, 45760, 46640, 47520, 48400, 49280, 50160, 51040, 51920, 52800, 53680, 54560, 55440, 56320, 57200, 58080, 58960, 59840, 60720, 61600, 62480, 63360, 64240, 65120, 66000, 66880, 67760, 68640, 69520, 70400, 71280, 72160, 73040, 73920, 74800, 75680, 76560, 77440, 78320, 79200, 80080, 80960, 81840, 82720, 83600, 84480, 85360, 86240, 87120, 88000, 88880, 89760, 90640, 91520, 92400, 93280, 94160, 95040, 95920, 96800, 97680, 98560, 99440

There is a total of 113 numbers (up to 100000) that are divisible by 880.
The sum of these numbers is 5668080.
The arithmetic mean of these numbers is 50160.

How to find the numbers divisible by 880?

Finding all the numbers that can be divided by 880 is essentially the same as searching for the multiples of 880: if a number N is a multiple of 880, then 880 is a divisor of N.

Indeed, if we assume that N is a multiple of 880, this means there exists an integer k such that:

$k \times 880 = N$

Conversely, the result of N divided by 880 is this same integer k (without any remainder):

$k = \frac{N}{880}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 880 (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 880 less than 100000):

1 × 880 = 880
2 × 880 = 1760
3 × 880 = 2640
...
112 × 880 = 98560
113 × 880 = 99440