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.


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