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Question Number 97752 by Power last updated on 09/Jun/20

Commented by mr W last updated on 09/Jun/20

$$igotx=1$$$$\Rightarrow AnswerB)$$

Commented by Power last updated on 09/Jun/20

$$\mathrm{sir}\mathrm{please}\mathrm{solution}$$

Commented by Algoritm last updated on 09/Jun/20

$$provethatsir$$

Commented by Farruxjano last updated on 09/Jun/20

$$\mathrm{I}\mathrm{got}\mathrm{either}.\mathrm{My}\mathrm{ans}\mathrm{is}\mathrm{x}=1\mathrm{too}.$$

Answered by Farruxjano last updated on 09/Jun/20

$$\mathrm{Theory}\mathrm{Ferma}:\mathrm{p}-\mathrm{prime}\mathrm{and}(\mathrm{a};\mathrm{p})=1$$$${\mathrm{a}}^{\mathrm{p}-1}\equiv 1\left(\mathrm{modp}\right){\mathrm{We}}^{\prime }\mathrm{ll}\mathrm{use}\mathrm{it}.$$$$1){10}^{{10}^{10}}\Rightarrow {10}^6\equiv 1\left(\mathrm{modp}\right){10}^{10}\equiv \mathrm{r}\left(\mathrm{mod}6\right)\Rightarrow$$$$\Rightarrow 2^9\cdot 5^{10}\equiv \frac{\mathrm{r}}{2}\left(\mathrm{mod}3\right)\Rightarrow \frac{\mathrm{r}}{2}=2\Rightarrow \mathrm{r}=4$$$$\Rightarrow {10}^{6\mathrm{k}+4}\equiv {10}^4\left(\mathrm{mod}7\right)\equiv 4\left(\mathrm{mod}7\right)\mathrm{r}=4$$$$2){20}^{{20}^{20}}={(7\cdot 3-1)}^{{20}^{20}}\equiv 1\left(\mathrm{mod}7\right)\Rightarrow \mathrm{r}=1$$$$3){30}^{{30}^{30}}={(4\cdot 7+2)}^{{30}^{30}}\equiv 2^{{30}^{30}}\left(\mathrm{mod}7\right)\Rightarrow$$$$2^6=64=9\cdot 7+1\equiv 1\left(\mathrm{mod}7\right)\Rightarrow {30}^{30}⋮6\Rightarrow$$$$\Rightarrow \mathrm{r}=1$$$$4){40}^{{40}^{40}}={(6\cdot 7-2)}^{{40}^{40}}\equiv 2^{{40}^{40}}\left(\mathrm{mod}7\right)\Rightarrow$$$${40}^{40}\equiv \mathrm{k}\left(\mathrm{mod}6\right)\Rightarrow 2^{119}\cdot 5^{40}\equiv \frac{\mathrm{k}}{2}\left(\mathrm{mod}3\right)\Rightarrow$$$$\frac{\mathrm{k}}{2}=2\Rightarrow \mathrm{k}=4\Rightarrow 2^{6\mathrm{k}+4}\equiv 2^4\left(\mathrm{mod}7\right)\equiv$$$$\equiv 2\left(\mathrm{mod}7\right)\Rightarrow \mathrm{r}=2$$$$\mathrm{So},\mathrm{in}\mathrm{general}:4+1+1+2=8\equiv 1\left(\mathrm{mod}7\right)$$$$\mathrm{Ans}:\mathit{R}=1$$

Commented by Algoritm last updated on 09/Jun/20

$$\mathrm{super}!$$

Commented by mr W last updated on 09/Jun/20

$$nicesolutionsir!$$

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