# What are the numbers divisible by 1001?

1001, 2002, 3003, 4004, 5005, 6006, 7007, 8008, 9009, 10010, 11011, 12012, 13013, 14014, 15015, 16016, 17017, 18018, 19019, 20020, 21021, 22022, 23023, 24024, 25025, 26026, 27027, 28028, 29029, 30030, 31031, 32032, 33033, 34034, 35035, 36036, 37037, 38038, 39039, 40040, 41041, 42042, 43043, 44044, 45045, 46046, 47047, 48048, 49049, 50050, 51051, 52052, 53053, 54054, 55055, 56056, 57057, 58058, 59059, 60060, 61061, 62062, 63063, 64064, 65065, 66066, 67067, 68068, 69069, 70070, 71071, 72072, 73073, 74074, 75075, 76076, 77077, 78078, 79079, 80080, 81081, 82082, 83083, 84084, 85085, 86086, 87087, 88088, 89089, 90090, 91091, 92092, 93093, 94094, 95095, 96096, 97097, 98098, 99099

- There is a
**total**of 99 numbers (up to 100000) that are divisible by 1001. - The
**sum**of these numbers is 4954950. - The arithmetic
**mean**of these numbers is 50050.

## How to find the numbers divisible by 1001?

Finding all the numbers that can be divided by 1001 is essentially the same as searching for the multiples of 1001: if a number `N` is a multiple of 1001, then 1001 is a divisor of `N`.

Indeed, if we assume that `N` is a multiple of 1001, this means there exists an integer `k` such that:

$k\times 1001=N$

Conversely, the result of `N` divided by 1001 is this same integer `k` (without any remainder):

$k=\frac{N}{\mathrm{1001}}$

From this we can see that, theoretically, there's an infinite quantity of multiples of 1001 (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 1001 less than 100000**):

- 1 × 1001 = 1001
- 2 × 1001 = 2002
- 3 × 1001 = 3003
- ...
- 98 × 1001 = 98098
- 99 × 1001 = 99099