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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.

natilie45632 Jun 16, 2021

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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

Guest · Answer 1 · 2021-06-16T07:18:06

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.