A positive integer is called a perfect power if it can be written in the form $a^{b}$, where $a$ and $b$ are positive integers with $b \geq 2$. For example, 32 and 125 are perfect powers because $32=2^{5}$ and $125=5^{3}$.

The increasing sequence

$$
2,3,5,6,7,10, \ldots
$$

consists of all positive integers which are not perfect powers. The sum of the squares of the digits of the 1000th number in this sequence is
(A) 42
(B) 26
(C) 33
(D) 18
(E) 21
Since we have an increasing sequence of integers, the 1000th term will be at least 1000 .

We start by determining the number of perfect powers less than or equal to 1000. (This will tell us how many integers less than 1000 are "skipped" by the sequence.)

Let us do this by making a list of all perfect powers less than 1100 (we will need to know the locations of some perfect powers bigger than 1000 anyways):

Perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, $324,361,400,441,484,529,576,625,676,729,784,841,900,961,1024,1089$

Perfect cubes: $1,8,27,64,125,216,343,512,729,1000$

Perfect 4th powers: $1,16,81,256,625$

Perfect 5th powers: 1, 32, 243, 1024

Perfect 6th powers: $1,64,729$

Perfect 7th powers: 1,128

Perfect 8th powers: 1,256

Perfect 9th powers: 1,512

Perfect 10th powers: 1,1024

In these lists, there are 41 distinct perfect powers less than or equal to 1000.

Thus, there are 959 positive integers less than or equal to 1000 which are not perfect powers. Therefore, 999 will be the 959th term in the sequence. (This is because 1000 itself is actually a perfect power; if it wasn't, 1000 would be the 959th term.)

Thus, 1001 will be the 960th term.

The next perfect power larger than 1000 is 1024 .

Thus, 1023 will be the 982 nd term and 1025 will be the 983 rd term.

The next perfect power larger than 1024 is 1089 .

Therefore, the 1042 will be the 1000th term.

The sum of the squares of the digits of 1042 is $1^{2}+0^{2}+4^{2}+2^{2}=21$.

ANsWER: (E)