# Question on series till 2009

Numbers 1, 2, 3 ……, 2009 are written in the natural order. Numbers in odd places are stricken off 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'.

• Better write them as $1, 10, 11, 100, \ldots, 11111011001$ – peterwhy Dec 27 '14 at 15:39

By writing numbers $1$ to $2009$ in binary form: $$1, 10, 11, 100, 101, 110, 111, 1000,1001 \ldots, 11111011001$$
The first round, numbers of last bit of $1$ are removed. The second round, numbers of the second last bit of $1$ are removed. Keep doing this, the last remaining number is the number in the form $100\ldots000_2$ between $1$ and $2009_{10}$ with the longest string of $0$'s, or $1024_{10}$.
Hint: In the first pass, all terms of the form $2n+1$ are struck off ($n\ge 0$), leaving only terms of the form $2n$. In the second pass, all terms of the form $2(2n+1) = 4n+2$ are struck off, leaving only terms of the form $4n$. And so on. How many passes are there, and what terms remain after the last pass?