The increasing sequence \(2, 3, 5, 6, 7, 10, \ldots\) consists of all positive integers which are not perfect powers. What is the sum of the squares of the digits of the 1000th number in this sequence?
(note that it says perfect powers, not just squares or cubes)
We start by listing out the perfect powers under 1100. There are 42, one of them being 1089 and the rest under or equal to 1024. Since the 1000th number of the sequence that does not have perfect powers taken out (and starts with 2) is 1001, we add 41 to obtain 1042. Since 1089 is not less than 1042, we can ignore the 1089. Therefore, the answer is 1^2+0^2+4^2+2^2=21.