Question

numbers 1, 2, 3....2009 are written in the natural number order . numbers in odd places are striken off from this sequence to obtain a new sequence . numbers in odd places are stricken off from this sequence to obtain another sequence and so on ....until only one term a is left ...then find a .....



Answer with EXPLANATION


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Answer 1

$\huge\bold\red{HOLA}$

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

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