What are the numbers divisible by 998?

998, 1996, 2994, 3992, 4990, 5988, 6986, 7984, 8982, 9980, 10978, 11976, 12974, 13972, 14970, 15968, 16966, 17964, 18962, 19960, 20958, 21956, 22954, 23952, 24950, 25948, 26946, 27944, 28942, 29940, 30938, 31936, 32934, 33932, 34930, 35928, 36926, 37924, 38922, 39920, 40918, 41916, 42914, 43912, 44910, 45908, 46906, 47904, 48902, 49900, 50898, 51896, 52894, 53892, 54890, 55888, 56886, 57884, 58882, 59880, 60878, 61876, 62874, 63872, 64870, 65868, 66866, 67864, 68862, 69860, 70858, 71856, 72854, 73852, 74850, 75848, 76846, 77844, 78842, 79840, 80838, 81836, 82834, 83832, 84830, 85828, 86826, 87824, 88822, 89820, 90818, 91816, 92814, 93812, 94810, 95808, 96806, 97804, 98802, 99800

There is a total of 100 numbers (up to 100000) that are divisible by 998.
The sum of these numbers is 5039900.
The arithmetic mean of these numbers is 50399.

How to find the numbers divisible by 998?

Finding all the numbers that can be divided by 998 is essentially the same as searching for the multiples of 998: if a number N is a multiple of 998, then 998 is a divisor of N.

Indeed, if we assume that N is a multiple of 998, this means there exists an integer k such that:

$k \times 998 = N$

Conversely, the result of N divided by 998 is this same integer k (without any remainder):

$k = \frac{N}{998}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 998 (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 998 less than 100000):

1 × 998 = 998
2 × 998 = 1996
3 × 998 = 2994
...
99 × 998 = 98802
100 × 998 = 99800