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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

$${Help} \\ $$

Answered by mahdipoor last updated on 12/Jan/22

N=44...488..88+1 =4×(((10^(2002) −1)/9))×10^(2002) +8×(((10^(2002) −1)/9))+1 (4/9)(10^(2002) −1)(10^(2002) +2)+1=(4/9)(10^(4004) +10^(2002) −2)+1= (1/9)(4×10^(4004) +4×10^(2002) +1)= (1/9)(2×10^(2002) +1)^2 ⇒(√N)=(1/3)(2×10^(2002) +1)=6(((10^(2002) −1)/9))+1= 66..667 (2002 times 6 repaet)

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

Commented by bekzodjumayev last updated on 12/Jan/22

Thanks

$${Thanks} \\ $$

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