Question

1000 people are standing in a queue. People standing at odd number in the queue are removed. For example 1,3,5,7,9 etc. Remaining people continue to stand in the queue. Now again people standing at odd numbers in the queue are removed. This process continues. By doing this the queue reduces ajd finally 9nly one persson is left. What is the position of this last person left in the queue?

Answer 1

Given :

Number of people standing in queue : 1000

People standing at odd number of places are removed each time.

To find :

Position of last lef person.

Solution :

It is given that 1000 people are standing in a queue,

Lets take there number as the position on which they are standing like,

the person standing in starting is 1 and the last is 1000 .

So as we remove the odd number people we will remove 1, 3, 5, 7, 9 , 2n-1 (where n is a natural number).

so the remaining people will be 2, 4 , 6, 8, 10, 2n (where n is a natural number).

we can see that only places which can be divided by 2 are remaining as we rearrange them as 500 remaining people as 1 , 2 ,3...500.

so when again we remove the odd place people then

only people remaining will be even number again which had the positions 4 8 12 16 20 ( multiple of 4 )

similarly we can see that now only 2² places are remaining so as we go on,

we can see that there is an arrangement if we remove n times like this only places remaining will be :

$\bf = {2}^{n} \: \: places$

So,

The biggest number less than 1000 which is nth power of 2 is 512.

as,

$\\ \bf \: {2}^{9} = 512$

So,

on removing 9 times only the odd places,

The last remaining place will be 512.