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How many positive integers n satisfy $\lfloor \sqrt[3]{n} \rfloor = 5$?

 
 Aug 1, 2022
 #2
avatar+2275 
+1

Only 3 does. 

 
 Aug 1, 2022
 #3
avatar+578 
+1

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

 
Vinculum  Aug 1, 2022
edited by Vinculum  Aug 1, 2022
 #4
avatar+2275 
+1

Oh, wait that's a cube root...

 

Yea, you're right, I thought it meant \(n\sqrt{3}\) not \(\sqrt[3]{n}\)

 
BuilderBoi  Aug 1, 2022
 #5
avatar+124525 
+2

\( $\lfloor \sqrt[3]{n} \rfloor = 5$\)

 

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

 

 

cool cool cool

 
 Aug 1, 2022

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