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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.

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up vote 5 down vote accepted

Hint: $$ [Z^k]=b\quad\Longleftrightarrow\quad b\leq Z^k<b+1 \quad\Longleftrightarrow\quad b^{1/k}\leq Z<(b+1)^{1/k} $$

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Please elaborate a little more, what to do after getting the 3 inequalities? – TheApe Aug 8 '12 at 21:31
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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 clear – Norbert Aug 8 '12 at 21:40
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19+27 = 46 Am I correct master? – TheApe Aug 8 '12 at 21:48
    
possible values are 1681, 1682, ..., 1727(total 46) – TheApe Aug 8 '12 at 21:48
    
hmm.. I got 47${}{}$ – Norbert Aug 8 '12 at 21:53

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