# Thread: Series

1. ## 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.

2. 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.

3. 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 .
So you only need to find 2^n<=1998

4. Thanks for the insight.

5. Originally Posted by darknight
Thanks for the insight.
No problem.

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