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Question Number 28932 by Rasheed.Sindhi last updated on 01/Feb/18
Determine the least number of 4 digits,  which is perfect square.  Method of finding is required.
$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{least}\:\mathrm{number}\:\mathrm{of}\:\mathrm{4}\:\mathrm{digits}, \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{perfect}\:\mathrm{square}. \\ $$$$\mathrm{Method}\:\mathrm{of}\:\mathrm{finding}\:\mathrm{is}\:\boldsymbol{\mathrm{required}}. \\ $$
Answered by mrW2 last updated on 01/Feb/18
x=n^2 ≥1000  n≥(√(1000))=31.6  ⇒n≥32  x_(min) =32^2 =1024    x=n^2 ≤9999  n≤(√(9999))=99.99  ⇒n≤99  x_(max) =99^2 =9801    32,33,...,99 ⇒ 99−32+1=68  i.e. there are totally 68  4−digit−numbers which are  perfect square.
$${x}={n}^{\mathrm{2}} \geqslant\mathrm{1000} \\ $$$${n}\geqslant\sqrt{\mathrm{1000}}=\mathrm{31}.\mathrm{6} \\ $$$$\Rightarrow{n}\geqslant\mathrm{32} \\ $$$${x}_{{min}} =\mathrm{32}^{\mathrm{2}} =\mathrm{1024} \\ $$$$ \\ $$$${x}={n}^{\mathrm{2}} \leqslant\mathrm{9999} \\ $$$${n}\leqslant\sqrt{\mathrm{9999}}=\mathrm{99}.\mathrm{99} \\ $$$$\Rightarrow{n}\leqslant\mathrm{99} \\ $$$${x}_{{max}} =\mathrm{99}^{\mathrm{2}} =\mathrm{9801} \\ $$$$ \\ $$$$\mathrm{32},\mathrm{33},…,\mathrm{99}\:\Rightarrow\:\mathrm{99}−\mathrm{32}+\mathrm{1}=\mathrm{68} \\ $$$${i}.{e}.\:{there}\:{are}\:{totally}\:\mathrm{68} \\ $$$$\mathrm{4}−{digit}−{numbers}\:{which}\:{are} \\ $$$${perfect}\:{square}. \\ $$
Commented by Rasheed.Sindhi last updated on 02/Feb/18
Thαnks-a-lot Sir!
$$\mathbb{T}\mathrm{h}\alpha\mathrm{nks}-\mathrm{a}-\mathrm{lot}\:\mathbb{S}\mathrm{ir}! \\ $$

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