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Question Number 164001 by bekzodjumayev last updated on 12/Jan/22

Commented by bekzodjumayev last updated on 12/Jan/22

$$Help$$

Answered by mahdipoor last updated on 12/Jan/22

$$N=44...488..88+1$$$$=4\times \left(\frac{{10}^{2002}-1}{9}\right)\times {10}^{2002}+8\times \left(\frac{{10}^{2002}-1}{9}\right)+1$$$$\frac{4}{9}({10}^{2002}-1)({10}^{2002}+2)+1=\frac{4}{9}({10}^{4004}+{10}^{2002}-2)+1=$$$$\frac{1}{9}(4\times {10}^{4004}+4\times {10}^{2002}+1)=$$$$\frac{1}{9}{(2\times {10}^{2002}+1)}^2$$$$\Rightarrow \sqrt{N}=\frac{1}{3}(2\times {10}^{2002}+1)=6\left(\frac{{10}^{2002}-1}{9}\right)+1=$$$$66..667\left(2002times6repaet\right)$$

Commented by bekzodjumayev last updated on 12/Jan/22

$$Thanks$$

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