How many positive integers n satisfy $\lfloor \sqrt[3]{n} \rfloor = 5$?
Only 3 does.
Hey BuilderBoi,
I have a question...
How is it 3?
Using the floor rule \(\mathrm{If}\:\lfloor \:n\:\rfloor =x\:\mathrm{then}\:x\le \:n \)
We get \(5\le \sqrt[3]{n}<6\)
That is \(125\le \:n<216\)
-Vinculum
Oh, wait that's a cube root...
Yea, you're right, I thought it meant \(n\sqrt{3}\) not \(\sqrt[3]{n}\)
\( $\lfloor \sqrt[3]{n} \rfloor = 5$\)
Note that (125)^(1/3) = 5
And that (216)^(1/3) = 6
So....the number of positive integers that satisfy this = 215 - 125 + 1 = 91
+2668 
