Let's continue the pattern and get all the
terms between 0 and 100.

1, 2, 4, 8, 16, 22, 24, 28, 36, 42, 44,
48, 56, 62, 64, 68, 76, 82, 84, 88, 96

Except for the very first term of 1, notice in passing that the 
pattern of last digits goes 2,4,8,6 and continues repeating that
sequence on and on. 

To get the next term after 96, we will add 6 to 96 and get 102, 
which ends in 02.  Therefore, except for the very first term of 01, 
this pattern of 20 pairs of last two digits continues on and on. 

02, 04, 08, 16, 22, 24, 28, 36, 42, 44,
48, 56, 62, 64, 68, 76, 82, 84, 88, 96

So except for the 21 terms between 0 and 100,
there are 20 between 100 and 200, 20 between
200 and 300, and so on up to 20 between 900,and 1000.

There are 10 groups of 100 less than 1000, and except
for the first term 1, there are 20 terms between each
two successive multiples of 100.  So the answer is 
20×10 = 200 plus the very first term 1, which does 
not fit the pattern.  Counting the 1 makes it 201.

Answer: 201.

Edwin