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 If⌊n⌋=xthenx≤n
We get 5≤3√n<6
That is 125≤n<216
-Vinculum
Oh, wait that's a cube root...
Yea, you're right, I thought it meant n√3 not 3√n
$⌊3√n⌋=5$
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 
