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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bubupin

2007-08-26 01:01
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