What are the numbers divisible by 828?

828, 1656, 2484, 3312, 4140, 4968, 5796, 6624, 7452, 8280, 9108, 9936, 10764, 11592, 12420, 13248, 14076, 14904, 15732, 16560, 17388, 18216, 19044, 19872, 20700, 21528, 22356, 23184, 24012, 24840, 25668, 26496, 27324, 28152, 28980, 29808, 30636, 31464, 32292, 33120, 33948, 34776, 35604, 36432, 37260, 38088, 38916, 39744, 40572, 41400, 42228, 43056, 43884, 44712, 45540, 46368, 47196, 48024, 48852, 49680, 50508, 51336, 52164, 52992, 53820, 54648, 55476, 56304, 57132, 57960, 58788, 59616, 60444, 61272, 62100, 62928, 63756, 64584, 65412, 66240, 67068, 67896, 68724, 69552, 70380, 71208, 72036, 72864, 73692, 74520, 75348, 76176, 77004, 77832, 78660, 79488, 80316, 81144, 81972, 82800, 83628, 84456, 85284, 86112, 86940, 87768, 88596, 89424, 90252, 91080, 91908, 92736, 93564, 94392, 95220, 96048, 96876, 97704, 98532, 99360

There is a total of 120 numbers (up to 100000) that are divisible by 828.
The sum of these numbers is 6011280.
The arithmetic mean of these numbers is 50094.

How to find the numbers divisible by 828?

Finding all the numbers that can be divided by 828 is essentially the same as searching for the multiples of 828: if a number N is a multiple of 828, then 828 is a divisor of N.

Indeed, if we assume that N is a multiple of 828, this means there exists an integer k such that:

$k \times 828 = N$

Conversely, the result of N divided by 828 is this same integer k (without any remainder):

$k = \frac{N}{828}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 828 (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 828 less than 100000):

1 × 828 = 828
2 × 828 = 1656
3 × 828 = 2484
...
119 × 828 = 98532
120 × 828 = 99360