What are the numbers divisible by 840?

840, 1680, 2520, 3360, 4200, 5040, 5880, 6720, 7560, 8400, 9240, 10080, 10920, 11760, 12600, 13440, 14280, 15120, 15960, 16800, 17640, 18480, 19320, 20160, 21000, 21840, 22680, 23520, 24360, 25200, 26040, 26880, 27720, 28560, 29400, 30240, 31080, 31920, 32760, 33600, 34440, 35280, 36120, 36960, 37800, 38640, 39480, 40320, 41160, 42000, 42840, 43680, 44520, 45360, 46200, 47040, 47880, 48720, 49560, 50400, 51240, 52080, 52920, 53760, 54600, 55440, 56280, 57120, 57960, 58800, 59640, 60480, 61320, 62160, 63000, 63840, 64680, 65520, 66360, 67200, 68040, 68880, 69720, 70560, 71400, 72240, 73080, 73920, 74760, 75600, 76440, 77280, 78120, 78960, 79800, 80640, 81480, 82320, 83160, 84000, 84840, 85680, 86520, 87360, 88200, 89040, 89880, 90720, 91560, 92400, 93240, 94080, 94920, 95760, 96600, 97440, 98280, 99120, 99960

There is a total of 119 numbers (up to 100000) that are divisible by 840.
The sum of these numbers is 5997600.
The arithmetic mean of these numbers is 50400.

How to find the numbers divisible by 840?

Finding all the numbers that can be divided by 840 is essentially the same as searching for the multiples of 840: if a number N is a multiple of 840, then 840 is a divisor of N.

Indeed, if we assume that N is a multiple of 840, this means there exists an integer k such that:

$k \times 840 = N$

Conversely, the result of N divided by 840 is this same integer k (without any remainder):

$k = \frac{N}{840}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 840 (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 840 less than 100000):

1 × 840 = 840
2 × 840 = 1680
3 × 840 = 2520
...
118 × 840 = 99120
119 × 840 = 99960