What are the numbers divisible by 963?

963, 1926, 2889, 3852, 4815, 5778, 6741, 7704, 8667, 9630, 10593, 11556, 12519, 13482, 14445, 15408, 16371, 17334, 18297, 19260, 20223, 21186, 22149, 23112, 24075, 25038, 26001, 26964, 27927, 28890, 29853, 30816, 31779, 32742, 33705, 34668, 35631, 36594, 37557, 38520, 39483, 40446, 41409, 42372, 43335, 44298, 45261, 46224, 47187, 48150, 49113, 50076, 51039, 52002, 52965, 53928, 54891, 55854, 56817, 57780, 58743, 59706, 60669, 61632, 62595, 63558, 64521, 65484, 66447, 67410, 68373, 69336, 70299, 71262, 72225, 73188, 74151, 75114, 76077, 77040, 78003, 78966, 79929, 80892, 81855, 82818, 83781, 84744, 85707, 86670, 87633, 88596, 89559, 90522, 91485, 92448, 93411, 94374, 95337, 96300, 97263, 98226, 99189

There is a total of 103 numbers (up to 100000) that are divisible by 963.
The sum of these numbers is 5157828.
The arithmetic mean of these numbers is 50076.

How to find the numbers divisible by 963?

Finding all the numbers that can be divided by 963 is essentially the same as searching for the multiples of 963: if a number N is a multiple of 963, then 963 is a divisor of N.

Indeed, if we assume that N is a multiple of 963, this means there exists an integer k such that:

$k \times 963 = N$

Conversely, the result of N divided by 963 is this same integer k (without any remainder):

$k = \frac{N}{963}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 963 (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 963 less than 100000):

1 × 963 = 963
2 × 963 = 1926
3 × 963 = 2889
...
102 × 963 = 98226
103 × 963 = 99189