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Find the smallest positive integer n ≥ 100 such that {√n} >1/2. 

Note: for a real number x, {x} = x − \(\lfloor x \rfloor\)denotes the fractional part of x.

 Jun 16, 2021

Best Answer 

 #1
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You can do this by trial and error. Since \(\sqrt{110 } \ is\ approximately\ 10.4880884817 \) with fractiona part equal to

                                                   

                                                        \(10.4880884817 -10 = 0.4880884817<0.5\),

 

and \(\sqrt{111}\) is approximately \(10.5356537529 \)with fractional part equal to

                                                        

                                                        \(10.5356537529 -10=0.5356537529>0.5\),

n = 111 must be the number you are looking for.

 Jun 16, 2021
 #1
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+1
Best Answer

You can do this by trial and error. Since \(\sqrt{110 } \ is\ approximately\ 10.4880884817 \) with fractiona part equal to

                                                   

                                                        \(10.4880884817 -10 = 0.4880884817<0.5\),

 

and \(\sqrt{111}\) is approximately \(10.5356537529 \)with fractional part equal to

                                                        

                                                        \(10.5356537529 -10=0.5356537529>0.5\),

n = 111 must be the number you are looking for.

Guest Jun 16, 2021

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