# Series

• Jun 8th 2010, 05:48 AM
darknight
Series
The numbers 1,2,3........1998 are written in natural order.
Numbers in odd places are stricken off to obtain a new sequence.
2,4,6,........
Then, numbers in odd places of this sequence are stricken off.
which will obtain: 4,8,......
This process continues until only one term 'a' is left. Then a = ?????

I started by,
2,4,6..........,1996, 1998
Using Arithmetic Progression:
1998 = 2 + (n-1)2
1996/2 = n-1
n = 998 + 1
n = 999

Therefore, in the next step 1998 will be stricken off
New sequence = 4,8,12......, 1992, 1996
Again, by A.P.
1996 = 4 + (n-1)4
1992/4 = n-1
n = 499
So, new sequence is:
8,16, 24..... 1984, 1992

So far:
2,4,6..........,1996, 1998
4,8,12......, 1992, 1996
8,16, 24..... 1984, 1992

I noticed that the differences were increasing multiples of 2.
1996 + 2 = 1998
1992 + 4 = 1996
And tried to deduce a, using arithmetic and geometric progressions but couldn't figure out how.

Since the number of terms in the sequence was exponentially decreasing, I used brute force.
I went on till the starting number of the sequence became 512:
and the sequences looked like this:
256, 512, 768,.... 1376
512, 1024... 1120

And completing the last iteration I got 'a' as 1024.

Is this correct? And secondly, isn't their a simpler way to solve this?
The calculation was long and tedious, and I worked hard, not smart.
Finally, this is a 2 mark question (out of a 60 mark test), so I'm sure there is a simpler way.
• Jun 8th 2010, 06:00 AM
undefined
Quote:

Originally Posted by darknight
Is this correct? And secondly, isn't their a simpler way to solve this?
The calculation was long and tedious, and I worked hard, not smart.
Finally, this is a 2 mark question (out of a 60 mark test), so I'm sure there is a simpler way.

It's correct. The answer will always be the greatest power of two less than or equal to the upper limit. I came to this conclusion from reviewing what you wrote. The reason is that it's impossible to eliminate any power of 2 from the list until it's the first element in the list.

A little more explanation: The first time you went through, you only removed odd numbers. The second time you went through, you only removed numbers that were divisible by 2 but not 4. Then you removed numbers divisible by 4 but not 8. Etc.
• Jun 8th 2010, 06:42 AM
p0oint
1,2,3,4,5,6,7,8,9,10,11,12.......1998.....
2,4,6,8,10,12,14,16,18,20...................
4,8,12,16,20,24,28,32,36....................
8................................................. ...
2^n............................................... .

You can think this way:
the first term actually will be the last term of the sequence after cutting the sequence n-1 times :D.
So you only need to find 2^n<=1998
• Jun 8th 2010, 06:45 AM
darknight
Thanks for the insight.
• Jun 8th 2010, 06:53 AM
p0oint
Quote:

Originally Posted by darknight
Thanks for the insight.

No problem.

Just check my last post (I have changed it), it is much neater solution.