please solve this interesting problem

suppose u r writing (2n+1) numbers starting from 1 i.e. 1,2,3,4,.....(2n+1) around a circle. You are now doing an operation that is as follows:::::you will delete the alternating numbers when you have come to the position where you have started writing the numbers i.e. ,think, u have written upto 17 and then comes the number 1 so keeping 1 as it is you delete the number 2,4,6,.... and when you delete 16 ,a cycle is covered and keep 17 as it is and then you will delete alternatingly the remaining numbers .you will find that after all operation there would be only one number left.find that particular number in general cases and in terms of "n" ...i told you an example taking (2n+1)=17 before...

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

Your procedure is unclear. In your example you keep 1 the first time around and the second time you keep the 17 which is not at the beginning the circular list. Please clarify.

Francisco Rivera - 8 years, 3 months ago

i am giving you a picture of the procedure my email is metronetizen@gmail.com please give your email to send that picture

sayan chowdhury - 8 years, 3 months ago

please please see the picture i uploaded as my picture ....you would get the idea

sayan chowdhury - 8 years, 3 months ago

This is the well-known Josephus Problem. As Wikipedia mentions, this is also important in computer science specifically as a classic example of "circular linked-list".

Peace. AD.

Abhishek De - 8 years, 3 months ago

i am giving you a picture of the procedure my email is metronetizen@gmail.com please give your email to send that picture

sayan chowdhury - 8 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$