What are the numbers divisible by 1005?
1005, 2010, 3015, 4020, 5025, 6030, 7035, 8040, 9045, 10050, 11055, 12060, 13065, 14070, 15075, 16080, 17085, 18090, 19095, 20100, 21105, 22110, 23115, 24120, 25125, 26130, 27135, 28140, 29145, 30150, 31155, 32160, 33165, 34170, 35175, 36180, 37185, 38190, 39195, 40200, 41205, 42210, 43215, 44220, 45225, 46230, 47235, 48240, 49245, 50250, 51255, 52260, 53265, 54270, 55275, 56280, 57285, 58290, 59295, 60300, 61305, 62310, 63315, 64320, 65325, 66330, 67335, 68340, 69345, 70350, 71355, 72360, 73365, 74370, 75375, 76380, 77385, 78390, 79395, 80400, 81405, 82410, 83415, 84420, 85425, 86430, 87435, 88440, 89445, 90450, 91455, 92460, 93465, 94470, 95475, 96480, 97485, 98490, 99495
- There is a total of 99 numbers (up to 100000) that are divisible by 1005.
- The sum of these numbers is 4974750.
- The arithmetic mean of these numbers is 50250.
How to find the numbers divisible by 1005?
Finding all the numbers that can be divided by 1005 is essentially the same as searching for the multiples of 1005: if a number N is a multiple of 1005, then 1005 is a divisor of N.
Indeed, if we assume that N is a multiple of 1005, this means there exists an integer k such that:
Conversely, the result of N divided by 1005 is this same integer k (without any remainder):
From this we can see that, theoretically, there's an infinite quantity of multiples of 1005 (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 1005 less than 100000):
- 1 × 1005 = 1005
- 2 × 1005 = 2010
- 3 × 1005 = 3015
- ...
- 98 × 1005 = 98490
- 99 × 1005 = 99495