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Question Number 130970 by naka3546 last updated on 31/Jan/21

$${2023}^{{2021}^{2019}}\mathrm{mod}1000=x$$$$\sqrt{\frac{x}{9}+\sqrt[8]{x+89}}=y$$$$x+y=?$$

Commented by AlagaIbile last updated on 31/Jan/21

$$Since\mathit{g}\mathit{c}\mathit{d}(2023,1000)=1.$$$$\lambda \left(1000\right)=100,\lambda \left(100\right)=20.\lambda (\cdot )denote$$$$CarmichealLambdaFunction$$$${2023}^{{2021}^{2019mod20}mod100}mod1000$$$$\equiv {23}^{{21}^{19}mod100}mod1000$$$$\equiv {23}^{81}mod1000$$$$\equiv {\left(167\right)}^{27}\equiv {\left(463\right)}^9\equiv {\left(847\right)}^{3}mod1000$$$$\mathit{x}\equiv 423mod1000$$$$\therefore \mathit{y}=\sqrt{\frac{423}{9}+\sqrt[8]{423+89}}=\sqrt{47+\sqrt[8]{512}}$$$$\mathit{x}+\mathit{y}=423+\sqrt{47+2\sqrt[8]{2}}$$$$\mathit{O}\mathit{b}\mathit{s}\mathit{e}\mathit{r}\mathit{v}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:Tomake\mathit{y}anintegerthequestionshouldaskfor$$$$\mathit{y}=\sqrt{\frac{\mathit{x}}{9}+\sqrt[9]{\mathit{x}+89}}=7$$$$\therefore \mathit{x}+\mathit{y}=423+7=450$$

Commented by naka3546 last updated on 31/Jan/21

$$Thankyou,sir.$$

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