Starting from a number, you can either cube it or divide it by 8. Starting from the number 2 and using these two operations, can you reach 64? And $2^{2011}$?
We can reach 64 since $64=\left(2^{3}\right)^{3} / 8$. However, we observe that after all the operations we end up with a power of 2 with an exponent that is a multiple of 3, and since 2011 is not divisible by 3, it is impossible to obtain $2^{2011}$.