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Question Number 187091 by Humble last updated on 13/Feb/23
⌊x^2 ⌋ − ⌊x⌋^2 =100 x∈R min(x)=?
$$ \\ $$$$\lfloor{x}^{\mathrm{2}} \rfloor\:\:−\:\:\lfloor{x}\rfloor^{\mathrm{2}} \:=\mathrm{100} \\ $$$${x}\in\mathbb{R} \\ $$$${min}\left({x}\right)=? \\ $$
Answered by MJS_new last updated on 13/Feb/23
x=n+f; n∈Z∧0≤f<1 ⌊x^2 ⌋−⌊x⌋^2 =100 ⌊n^2 +2nf+f^2 ⌋−⌊n+f⌋^2 =100 n^2 +⌊2nf+f^2 ⌋−n^2 =100 ⌊2nf+f^2 ⌋=100 2nf+f^2 −100=0∧0≤f<1∧n∈Z ⇒ f=−n+(√(n^2 +100)) 0≤−n+(√(n^2 +100))<1 ⇒ n≥50 n=50 ⇒ f=−50+10(√(26)) x=n+f=10(√(26))
$${x}={n}+{f};\:{n}\in\mathbb{Z}\wedge\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$$\lfloor{x}^{\mathrm{2}} \rfloor−\lfloor{x}\rfloor^{\mathrm{2}} =\mathrm{100} \\ $$$$\lfloor{n}^{\mathrm{2}} +\mathrm{2}{nf}+{f}^{\mathrm{2}} \rfloor−\lfloor{n}+{f}\rfloor^{\mathrm{2}} =\mathrm{100} \\ $$$${n}^{\mathrm{2}} +\lfloor\mathrm{2}{nf}+{f}^{\mathrm{2}} \rfloor−{n}^{\mathrm{2}} =\mathrm{100} \\ $$$$\lfloor\mathrm{2}{nf}+{f}^{\mathrm{2}} \rfloor=\mathrm{100} \\ $$$$\mathrm{2}{nf}+{f}^{\mathrm{2}} −\mathrm{100}=\mathrm{0}\wedge\mathrm{0}\leqslant{f}<\mathrm{1}\wedge{n}\in\mathbb{Z} \\ $$$$\Rightarrow\:{f}=−{n}+\sqrt{{n}^{\mathrm{2}} +\mathrm{100}} \\ $$$$\mathrm{0}\leqslant−{n}+\sqrt{{n}^{\mathrm{2}} +\mathrm{100}}<\mathrm{1} \\ $$$$\Rightarrow\:{n}\geqslant\mathrm{50} \\ $$$${n}=\mathrm{50}\:\Rightarrow\:{f}=−\mathrm{50}+\mathrm{10}\sqrt{\mathrm{26}} \\ $$$${x}={n}+{f}=\mathrm{10}\sqrt{\mathrm{26}} \\ $$
Commented by Humble last updated on 13/Feb/23
thank you, sir
$${thank}\:{you},\:{sir} \\ $$
Commented by MJS_new last updated on 13/Feb/23
you′re welcome
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$
Answered by mr W last updated on 13/Feb/23
x=n+f ⌊n^2 +2nf+f^2 ⌋−⌊n+f⌋^2 =100 n^2 +⌊2nf+f^2 ⌋−n^2 =100 100=⌊2nf+f^2 ⌋<2n+1 ⇒n>((99)/2) ⇒n≥50 100≤2nf+f^2 <101 f^2 +2nf−100≥0 ⇒f≥−n+(√(n^2 +100)) f^2 +2nf−101<0 ⇒f<−n+(√(n^2 +101)) general solution: (√(n^2 +100))≤x<(√(n^2 +101)) (n≥50) min(x)=(√(50^2 +100))=10(√(26))
$${x}={n}+{f} \\ $$$$\lfloor{n}^{\mathrm{2}} +\mathrm{2}{nf}+{f}^{\mathrm{2}} \rfloor−\lfloor{n}+{f}\rfloor^{\mathrm{2}} =\mathrm{100} \\ $$$${n}^{\mathrm{2}} +\lfloor\mathrm{2}{nf}+{f}^{\mathrm{2}} \rfloor−{n}^{\mathrm{2}} =\mathrm{100} \\ $$$$\mathrm{100}=\lfloor\mathrm{2}{nf}+{f}^{\mathrm{2}} \rfloor<\mathrm{2}{n}+\mathrm{1}\:\Rightarrow{n}>\frac{\mathrm{99}}{\mathrm{2}}\:\Rightarrow{n}\geqslant\mathrm{50} \\ $$$$\mathrm{100}\leqslant\mathrm{2}{nf}+{f}^{\mathrm{2}} <\mathrm{101} \\ $$$${f}^{\mathrm{2}} +\mathrm{2}{nf}−\mathrm{100}\geqslant\mathrm{0} \\ $$$$\Rightarrow{f}\geqslant−{n}+\sqrt{{n}^{\mathrm{2}} +\mathrm{100}} \\ $$$${f}^{\mathrm{2}} +\mathrm{2}{nf}−\mathrm{101}<\mathrm{0} \\ $$$$\Rightarrow{f}<−{n}+\sqrt{{n}^{\mathrm{2}} +\mathrm{101}} \\ $$$${general}\:{solution}: \\ $$$$\sqrt{{n}^{\mathrm{2}} +\mathrm{100}}\leqslant{x}<\sqrt{{n}^{\mathrm{2}} +\mathrm{101}} \\ $$$$\left({n}\geqslant\mathrm{50}\right) \\ $$$${min}\left({x}\right)=\sqrt{\mathrm{50}^{\mathrm{2}} +\mathrm{100}}=\mathrm{10}\sqrt{\mathrm{26}} \\ $$
Commented by Humble last updated on 13/Feb/23
thank you, Boss.
$${thank}\:{you},\:{Boss}. \\ $$
Commented by mr W last updated on 13/Feb/23
you too!
$${you}\:{too}! \\ $$

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