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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.

1 Answers 1

6

Hint: $$ [Z^k]=b\quad\Longleftrightarrow\quad b\leq Z^k

  • 0
    Please elaborate a little more, what to do after getting the 3 inequalities?2012-08-08
  • 3
    You should intersect this three intervals, then you'll get a zet of all possible $Z$'s which satisfy your equlitites. Given this intersection you can find range of values of $Z^6$. I think the rest is clear2012-08-08
  • 1
    19+27 = 46 Am I correct master?2012-08-08
  • 0
    possible values are 1681, 1682, ..., 1727(total 46)2012-08-08
  • 1
    hmm.. I got 47${}{}$2012-08-08
  • 0
    sorry, my bad :(2012-08-08