Its given that $$[Z]=3$$ $$[Z^{2}]=11$$ $$[Z^{3}]=41$$
Then, what is the number of all possible values of $[Z^{6}]$ where $[\;\cdot\;]$ is floor function.
Its given that $$[Z]=3$$ $$[Z^{2}]=11$$ $$[Z^{3}]=41$$
Then, what is the number of all possible values of $[Z^{6}]$ where $[\;\cdot\;]$ is floor function.
Hint: $$ [Z^k]=b\quad\Longleftrightarrow\quad b\leq Z^k