Question:
The sequence 2, 3, 5, 6, 7, 10, 11, $\ldots$ contains all the positive integers from least to greatest that are neither squares nor cubes nor perfect fifth powers (in the form of $x^{5}$, where $x$ is an integer). What is the $1000^{\mathrm{th}}$ term of the sequence?

Answer:
We see that it's easier to count the number of perfect squares, perfect cubes and perfect fifth powers less than $33^{2}=1089$. We see there are 32 perfect squares less than 1089, which are $1^2$, $2^2$, $\ldots$, $32^2$ and then there are 10 perfect cubes which are $1^3$, $\ldots$, $10^3$. There are 4 perfect fifth powers less than 1089 which are $1^5$, $\ldots$, $4^5$. Then notice there are 3 numbers that are both perfect squares and perfect cubes which are 1, $2^{6} = 64$ and $3^{6} = 729$. There are also 2 numbers that are both perfect squares and perfect fifth powers which are $1^{10} = 1$ and $2^{10} = 1024$. The only number which is both a perfect cube and a perfect fifth power is $1^{15}=1$. The only number which is a perfect square, perfect cube, and perfect fifth power all at the same time is $1^{30}=1$. So within the first 1089 positive integers we need to get rid of $32+10+4-3-2-1+1 =41$ integers which means the $1000^{\text{th}}$ term is $1000+41 = \boxed{1041}$.