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.