| 發布者 | 內容列 | learing
Not too shy to talk
註冊日: 2007-03-21
發表數: 39
|  question |  | From the natural number 1 2 3 4 5......, delete the multiples of 2 and the multiples of 3, but keep the multiples of 5. the remaining numbers are:
1,5,7,10,11,13,15,17,19,20,23,25,29,20......
In this sequence, what is the 2006th number?
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數學=樂趣
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| | 2007-08-25 15:25 |  | bubupin
Home away from home
註冊日: 2007-03-13
發表數: 353
| | Re: question | | A0={1,5,7,10,11,13,15,17,19,20,23,25,29,30}
A1={31,35,37,40,41,43,45,47,49,50,53,55,59,60}
.....................................................................
Ak={30*k+1，30*k+5，............，30*k+30}
k=[n/14]
[2006/14]=143，mod(2006,14)=4
The 2006th number=30*143+10=4300
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| | 2007-08-26 01:01 | |
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