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