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Question Number 196388 by sonukgindia last updated on 24/Aug/23

Answered by deleteduser1 last updated on 24/Aug/23

φ(20)=φ(4)φ(5)=(2^2 −2)(5−1)=8

$$\phi\left(\mathrm{20}\right)=\phi\left(\mathrm{4}\right)\phi\left(\mathrm{5}\right)=\left(\mathrm{2}^{\mathrm{2}} −\mathrm{2}\right)\left(\mathrm{5}−\mathrm{1}\right)=\mathrm{8} \\ $$

Answered by BaliramKumar last updated on 24/Aug/23

φ(20) = φ(2^2 ×5^1 ) = (2^2 −2^1 )(5^1 −5^0 ) = 8     determinant (((φ(P^n ) = P^n −P^(n−1) )))

$$\phi\left(\mathrm{20}\right)\:=\:\phi\left(\mathrm{2}^{\mathrm{2}} ×\mathrm{5}^{\mathrm{1}} \right)\:=\:\left(\mathrm{2}^{\mathrm{2}} −\mathrm{2}^{\mathrm{1}} \right)\left(\mathrm{5}^{\mathrm{1}} −\mathrm{5}^{\mathrm{0}} \right)\:=\:\mathrm{8} \\ $$$$ \\ $$$$\begin{array}{|c|}{\phi\left(\mathrm{P}^{\mathrm{n}} \right)\:=\:\mathrm{P}^{\mathrm{n}} −\mathrm{P}^{\mathrm{n}−\mathrm{1}} }\\\hline\end{array} \\ $$

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