What are the numbers divisible by 1000?

1000, 2000, 3000, 4000, 5000, 6000, 7000, 8000, 9000, 10000, 11000, 12000, 13000, 14000, 15000, 16000, 17000, 18000, 19000, 20000, 21000, 22000, 23000, 24000, 25000, 26000, 27000, 28000, 29000, 30000, 31000, 32000, 33000, 34000, 35000, 36000, 37000, 38000, 39000, 40000, 41000, 42000, 43000, 44000, 45000, 46000, 47000, 48000, 49000, 50000, 51000, 52000, 53000, 54000, 55000, 56000, 57000, 58000, 59000, 60000, 61000, 62000, 63000, 64000, 65000, 66000, 67000, 68000, 69000, 70000, 71000, 72000, 73000, 74000, 75000, 76000, 77000, 78000, 79000, 80000, 81000, 82000, 83000, 84000, 85000, 86000, 87000, 88000, 89000, 90000, 91000, 92000, 93000, 94000, 95000, 96000, 97000, 98000, 99000, 100000

There is a total of 100 numbers (up to 100000) that are divisible by 1000.
The sum of these numbers is 5050000.
The arithmetic mean of these numbers is 50500.

How to find the numbers divisible by 1000?

Finding all the numbers that can be divided by 1000 is essentially the same as searching for the multiples of 1000: if a number N is a multiple of 1000, then 1000 is a divisor of N.

Indeed, if we assume that N is a multiple of 1000, this means there exists an integer k such that:

$k \times 1000 = N$

Conversely, the result of N divided by 1000 is this same integer k (without any remainder):

$k = \frac{N}{1000}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 1000 (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 1000 less than 100000):

1 × 1000 = 1000
2 × 1000 = 2000
3 × 1000 = 3000
...
99 × 1000 = 99000
100 × 1000 = 100000