Question Number 56719 by naka3546 last updated on 22/Mar/19
$${The}\:\:{possible}\:\:{value}\:\:{of}\:\:{x}\:\:{that}\:\:{satisfy}\:\:: \\ $$$$\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} \:+\:\lfloor{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\rfloor\:+\:\lceil{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{1}\rceil\:\:=\:\:\mathrm{2017}\:\:,\:\:{x}\:\in\:\:\mathbb{R}^{+} \\ $$$${and}\:\:\:\lfloor{x}^{\mathrm{2}} \rfloor\:−\:\lceil\mathrm{2}{x}\rceil\:\:=\:\:... \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 22/Mar/19
Commented by tanmay.chaudhury50@gmail.com last updated on 22/Mar/19
Commented by tanmay.chaudhury50@gmail.com last updated on 22/Mar/19
Commented by tanmay.chaudhury50@gmail.com last updated on 22/Mar/19
$${x}=\mathrm{25}.\mathrm{923} \\ $$$$\lfloor\left(\mathrm{25}.\mathrm{923}\right)^{\mathrm{2}} \rfloor−\lceil\mathrm{2}×\mathrm{25}.\mathrm{923}\rceil \\ $$$$=\lfloor\mathrm{672}.\mathrm{001929}\rfloor−\lceil\mathrm{51}.\mathrm{846}\rceil \\ $$$$=\mathrm{672}−\mathrm{52} \\ $$$$=\mathrm{620} \\ $$
Commented by naka3546 last updated on 22/Mar/19
$${Sir},\:{the}\:\:{answer}\:\:{x}\:\:{not}\:\:{suitable}\:\:{for}\:\:{the}\:\:{equation}\:\:{above}\:. \\ $$$${Please},\:\:{recheck}\:\:{the}\:\:{answer}\:. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 22/Mar/19
$${ok}\:\:\:{let}\:{me}\:{check}... \\ $$
Answered by mr W last updated on 23/Mar/19
$${x}^{\mathrm{2}} \:+\:\lfloor{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\rfloor\:+\:\lceil{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{1}\rceil\:\:=\:\:\mathrm{2017} \\ $$$${x}^{\mathrm{2}} +{integer}+{integer}={integer} \\ $$$$\Rightarrow{x}^{\mathrm{2}} ={integer},\:{say}\:{x}^{\mathrm{2}} ={n} \\ $$$${n}+{n}+\lfloor−\mathrm{2}\sqrt{{n}}\rfloor+{n}+\lceil\mathrm{2}\sqrt{{n}}\rceil+\mathrm{1}=\mathrm{2017} \\ $$$${since}\:\lfloor−{x}\rfloor=−\lceil{x}\rceil \\ $$$$\Rightarrow\lfloor−\mathrm{2}\sqrt{{n}}\rfloor+\lceil\mathrm{2}\sqrt{{n}}\rceil=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{n}=\mathrm{2016} \\ $$$$\Rightarrow{n}=\mathrm{672} \\ $$$$\Rightarrow{x}=\sqrt{{n}}=\sqrt{\mathrm{672}}=\mathrm{4}\sqrt{\mathrm{42}}\approx\mathrm{25}.\mathrm{923} \\ $$$$\lfloor{x}^{\mathrm{2}} \rfloor−\lceil\mathrm{2}{x}\rceil=\mathrm{672}−\mathrm{52}=\mathrm{620} \\ $$
Commented by naka3546 last updated on 23/Mar/19
$${but},\:\:{if}\:\:{x}\:\:{is}\:\:{substituted}\:\:{to}\:\:{equation}\:\:{above}\:. \\ $$$${Is}\:\:{x}\:\:{right}\:\:?\:\: \\ $$$${x}^{\mathrm{2}} \:+\:\lfloor{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\rfloor\:+\:\lceil{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{1}\rceil\:\:=\:\:\mathrm{2015}\:\: \\ $$$${how}\:\:{is}\:\:{it}\:\:{happen}\:? \\ $$
Commented by mr W last updated on 23/Mar/19
$${how}\:{did}\:{you}\:{get}\:\mathrm{2015}?\:{this}\:{is}\:{how}\:{it}\:{is}: \\ $$$${x}^{\mathrm{2}} =\mathrm{672} \\ $$$${x}=\sqrt{\mathrm{672}}\approx\mathrm{25}.\mathrm{92} \\ $$$$\mathrm{2}{x}\approx\mathrm{51}.\mathrm{8} \\ $$$$\lfloor{x}^{\mathrm{2}} −\mathrm{2}{x}\rfloor=\lfloor\mathrm{672}−\mathrm{51}.\mathrm{8}\rfloor=\lfloor\mathrm{620}.\mathrm{2}\rfloor=\mathrm{620} \\ $$$$\lceil{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\rceil=\lceil\mathrm{672}+\mathrm{51}.\mathrm{8}+\mathrm{1}\rceil=\lceil\mathrm{724}.\mathrm{8}\rceil=\mathrm{725} \\ $$$${x}^{\mathrm{2}} +\lfloor{x}^{\mathrm{2}} −\mathrm{2}{x}\rfloor+\lceil{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\rceil=\mathrm{672}+\mathrm{620}+\mathrm{725}=\mathrm{2017} \\ $$$${so}\:{x}=\sqrt{\mathrm{672}}\:{is}\:{correct}. \\ $$