Question Number 143072 by Feruzbek last updated on 09/Jun/21
Answered by mr W last updated on 09/Jun/21
$${say}\:{x}={n}+{f} \\ $$$$\mathrm{10000}{n}=\left({n}+{f}\right)^{\mathrm{2}} \geqslant{n}^{\mathrm{2}} \\ $$$${n}^{\mathrm{2}} −\mathrm{10000}{n}\leqslant\mathrm{0}\: \\ $$$$\Rightarrow\mathrm{0}\leqslant{n}\leqslant\mathrm{10000}\:\:\:...\left({i}\right) \\ $$$$ \\ $$$$\mathrm{10000}{n}=\left({n}+{f}\right)^{\mathrm{2}} <\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${n}^{\mathrm{2}} −\mathrm{9998}{n}+\mathrm{1}>\mathrm{0} \\ $$$$\Rightarrow{n}<\mathrm{1}\:{or}\:{n}>\mathrm{9997}\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$$\Rightarrow{n}=\mathrm{0}\:{or}\:\mathrm{9998}\:{or}\:\mathrm{9999}\:{or}\:\mathrm{10000} \\ $$$$\Rightarrow{x}=\mathrm{100}\sqrt{{n}} \\ $$$$\:{i}.{e}.\:{x}\:=\mathrm{0}\:{or} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{100}\sqrt{\mathrm{9998}}\:{or} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{100}\sqrt{\mathrm{9999}}=\mathrm{300}\sqrt{\mathrm{1111}}\:{or} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{10000} \\ $$