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

 Apr 6, 2020
 #1
avatar+111329 
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\( \lfloor \sqrt{n} \rfloor = 5 \)

 

Note  that

 

36  = 6^2

So

floor √35 =  5

 

And note  that

25 = 5^2

So

floor √25 =  5

 

So....the number of positive integers, n, that satisfy  this  =  35 - 25 + 1  =   11

 

 

cool cool cool

 Apr 7, 2020

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