Numbers 1, 2, 3 ……, 2009 are written in the natural order. Numbers in oddplaces are stricken off to obtain a new sequence. Numbers in odd places arestricken off from this sequence to obtain another sequence and so on, until onlyone term a is left. Then find a

The answer of this question is 1024.

Solution: The no. are 1 , 2 , 3 , 4 , 5 , 6 , 7 ... 2009

If the no. odd places are sticken off , then the new sequence formed is

2, 4, 6, 8, ... 2008          -----------> no of form 2n where n can take form from 1,2...

If the no. odd places are sticken off , then the new sequence formed is

4, 8, 12, 16, ... 2008          -----------> no of form 4n where n can take form from 1,2...

If the no. odd places are sticken off , then the new sequence formed is

8, 16, 24 , ... 2008          -----------> no of form 8n where n can take form from 1,2...

If the no. odd places are sticken off , then the new sequence formed is

16, 32 , 48 ...          -----------> no of form 16n where n can take form from 1,2...

This process will go so on .

Then the sequences obatained will be n , 2n , 4n , 8n, 16n , 32n ...

Until the sequence of form 1024n wll be formed.further sequnces are not possible as it will be 2048n which is >2009

thus the last term is 1024(1) = 1024