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Question-97752




Question Number 97752 by Power last updated on 09/Jun/20
Commented by mr W last updated on 09/Jun/20
i got x=1  ⇒Answer B)
$${i}\:{got}\:{x}=\mathrm{1} \\ $$$$\left.\Rightarrow{Answer}\:{B}\right) \\ $$
Commented by Power last updated on 09/Jun/20
sir please solution
$$\mathrm{sir}\:\mathrm{please}\:\mathrm{solution} \\ $$
Commented by Algoritm last updated on 09/Jun/20
prove that sir
$${prove}\:{that}\:{sir}\: \\ $$
Commented by Farruxjano last updated on 09/Jun/20
I got either. My ans is x=1 too.
$$\mathrm{I}\:\mathrm{got}\:\mathrm{either}.\:\mathrm{My}\:\mathrm{ans}\:\mathrm{is}\:\mathrm{x}=\mathrm{1}\:\mathrm{too}. \\ $$
Answered by Farruxjano last updated on 09/Jun/20
Theory Ferma: p−prime and (a;p)=1  a^(p−1) ≡1(modp)   We′ll use it.  1)10^(10^(10) )  ⇒ 10^6 ≡1(modp)  10^(10) ≡r(mod6)⇒  ⇒ 2^9 ∙5^(10) ≡(r/2)(mod3) ⇒ (r/2)=2 ⇒ r=4  ⇒ 10^(6k+4) ≡10^4 (mod7)≡4(mod7) r=4  2)20^(20^(20) ) =(7∙3−1)^(20^(20) ) ≡1(mod7) ⇒ r=1  3)30^(30^(30) ) =(4∙7+2)^(30^(30) ) ≡2^(30^(30) ) (mod7) ⇒  2^6 =64=9∙7+1≡1(mod7) ⇒ 30^(30) ⋮6 ⇒  ⇒ r=1  4) 40^(40^(40) ) =(6∙7−2)^(40^(40) ) ≡2^(40^(40) ) (mod7) ⇒  40^(40) ≡k(mod6) ⇒ 2^(119) ∙5^(40) ≡(k/2)(mod3) ⇒  (k/2)=2 ⇒ k=4 ⇒ 2^(6k+4) ≡2^4 (mod7)≡  ≡2(mod7) ⇒ r=2  So, in general: 4+1+1+2=8≡1(mod7)  Ans: R=1
$$\mathrm{Theory}\:\mathrm{Ferma}:\:\mathrm{p}−\mathrm{prime}\:\mathrm{and}\:\left(\mathrm{a};\mathrm{p}\right)=\mathrm{1} \\ $$$$\mathrm{a}^{\mathrm{p}−\mathrm{1}} \equiv\mathrm{1}\left(\mathrm{modp}\right)\:\:\:\mathrm{We}'\mathrm{ll}\:\mathrm{use}\:\mathrm{it}. \\ $$$$\left.\mathrm{1}\right)\mathrm{10}^{\mathrm{10}^{\mathrm{10}} } \:\Rightarrow\:\mathrm{10}^{\mathrm{6}} \equiv\mathrm{1}\left(\mathrm{modp}\right)\:\:\mathrm{10}^{\mathrm{10}} \equiv\mathrm{r}\left(\mathrm{mod6}\right)\Rightarrow \\ $$$$\Rightarrow\:\mathrm{2}^{\mathrm{9}} \centerdot\mathrm{5}^{\mathrm{10}} \equiv\frac{\mathrm{r}}{\mathrm{2}}\left(\mathrm{mod3}\right)\:\Rightarrow\:\frac{\mathrm{r}}{\mathrm{2}}=\mathrm{2}\:\Rightarrow\:\mathrm{r}=\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{10}^{\mathrm{6k}+\mathrm{4}} \equiv\mathrm{10}^{\mathrm{4}} \left(\mathrm{mod7}\right)\equiv\mathrm{4}\left(\mathrm{mod7}\right)\:\mathrm{r}=\mathrm{4} \\ $$$$\left.\mathrm{2}\right)\mathrm{20}^{\mathrm{20}^{\mathrm{20}} } =\left(\mathrm{7}\centerdot\mathrm{3}−\mathrm{1}\right)^{\mathrm{20}^{\mathrm{20}} } \equiv\mathrm{1}\left(\mathrm{mod7}\right)\:\Rightarrow\:\mathrm{r}=\mathrm{1} \\ $$$$\left.\mathrm{3}\right)\mathrm{30}^{\mathrm{30}^{\mathrm{30}} } =\left(\mathrm{4}\centerdot\mathrm{7}+\mathrm{2}\right)^{\mathrm{30}^{\mathrm{30}} } \equiv\mathrm{2}^{\mathrm{30}^{\mathrm{30}} } \left(\mathrm{mod7}\right)\:\Rightarrow \\ $$$$\mathrm{2}^{\mathrm{6}} =\mathrm{64}=\mathrm{9}\centerdot\mathrm{7}+\mathrm{1}\equiv\mathrm{1}\left(\mathrm{mod7}\right)\:\Rightarrow\:\mathrm{30}^{\mathrm{30}} \vdots\mathrm{6}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{r}=\mathrm{1} \\ $$$$\left.\mathrm{4}\right)\:\mathrm{40}^{\mathrm{40}^{\mathrm{40}} } =\left(\mathrm{6}\centerdot\mathrm{7}−\mathrm{2}\right)^{\mathrm{40}^{\mathrm{40}} } \equiv\mathrm{2}^{\mathrm{40}^{\mathrm{40}} } \left(\mathrm{mod7}\right)\:\Rightarrow \\ $$$$\mathrm{40}^{\mathrm{40}} \equiv\mathrm{k}\left(\mathrm{mod6}\right)\:\Rightarrow\:\mathrm{2}^{\mathrm{119}} \centerdot\mathrm{5}^{\mathrm{40}} \equiv\frac{\mathrm{k}}{\mathrm{2}}\left(\mathrm{mod3}\right)\:\Rightarrow \\ $$$$\frac{\mathrm{k}}{\mathrm{2}}=\mathrm{2}\:\Rightarrow\:\mathrm{k}=\mathrm{4}\:\Rightarrow\:\mathrm{2}^{\mathrm{6k}+\mathrm{4}} \equiv\mathrm{2}^{\mathrm{4}} \left(\mathrm{mod7}\right)\equiv \\ $$$$\equiv\mathrm{2}\left(\mathrm{mod7}\right)\:\Rightarrow\:\mathrm{r}=\mathrm{2} \\ $$$$\mathrm{So},\:\mathrm{in}\:\mathrm{general}:\:\mathrm{4}+\mathrm{1}+\mathrm{1}+\mathrm{2}=\mathrm{8}\equiv\mathrm{1}\left(\mathrm{mod7}\right) \\ $$$$\mathrm{Ans}:\:\boldsymbol{{R}}=\mathrm{1} \\ $$
Commented by Algoritm last updated on 09/Jun/20
super!
$$\mathrm{super}! \\ $$
Commented by mr W last updated on 09/Jun/20
nice solution sir!
$${nice}\:{solution}\:{sir}! \\ $$

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