Menu Close

Question-96671




Question Number 96671 by 175 last updated on 03/Jun/20
Answered by maths mind last updated on 03/Jun/20
n=Π_(i=1) ^r p_i ^a_i  ⇒  ϕ(n)=Π_(i=1) ^r (p_i −1)p_i ^(a_i −1) =16=2^4   p_1 <p_2 <p_3 .....<p_r ⇒r≤3& p_r ≤17  r=1⇒ϕ(n)=(p_1 −1)p_1 ^(a_1 −1) =16⇒p_1 =2 ,a_1 =5⇔n=32  r=2⇒(p_1 −1)p_1 ^(a_1 −1) .(p_2 −1).p_2 ^(a_2 −1) =16⇒  p_2 =17⇒p_1 =2&a_1 =1  n=34,p_2 =5⇒a_2 =1,⇒(p_1 −1).p_1 ^(a_1 −1) =4  p_1 =2,a_1 =3⇒n=2^3 .5=40  p_2 =3⇒a_2 =1⇒p_1 =1&a_1 =4⇒n=2^4 .3=48  r=3⇒  (p_1 −1)(p_2 −1)(p_3 −1)≤16⇒  (p_1 −1)(p_2 −1)<4⇒p_2 ≤5⇒p_2 ∈{3,5}  p_3 ≤9⇒p_3 ∈{2,3,5,7},p_3 >p_2 ⇒p_3 ∈{5,7}⇒p_3 =5&p_2 =3  ⇒(p_1 −1)p_1 ^(a_1 −1) 2.3^(a_2 −1) .4.5^(a_3 −1) =16⇒  a_2 =a_3 =1&p_1 =2&a_1 =2⇒n=4.3.5=60
$${n}=\underset{{i}=\mathrm{1}} {\overset{{r}} {\prod}}{p}_{{i}} ^{{a}_{{i}} } \Rightarrow \\ $$$$\varphi\left({n}\right)=\underset{{i}=\mathrm{1}} {\overset{{r}} {\prod}}\left({p}_{{i}} −\mathrm{1}\right){p}_{{i}} ^{{a}_{{i}} −\mathrm{1}} =\mathrm{16}=\mathrm{2}^{\mathrm{4}} \\ $$$${p}_{\mathrm{1}} <{p}_{\mathrm{2}} <{p}_{\mathrm{3}} …..<{p}_{{r}} \Rightarrow{r}\leqslant\mathrm{3\&}\:{p}_{{r}} \leqslant\mathrm{17} \\ $$$${r}=\mathrm{1}\Rightarrow\varphi\left({n}\right)=\left({p}_{\mathrm{1}} −\mathrm{1}\right){p}_{\mathrm{1}} ^{{a}_{\mathrm{1}} −\mathrm{1}} =\mathrm{16}\Rightarrow{p}_{\mathrm{1}} =\mathrm{2}\:,{a}_{\mathrm{1}} =\mathrm{5}\Leftrightarrow{n}=\mathrm{32} \\ $$$${r}=\mathrm{2}\Rightarrow\left({p}_{\mathrm{1}} −\mathrm{1}\right){p}_{\mathrm{1}} ^{{a}_{\mathrm{1}} −\mathrm{1}} .\left({p}_{\mathrm{2}} −\mathrm{1}\right).{p}_{\mathrm{2}} ^{{a}_{\mathrm{2}} −\mathrm{1}} =\mathrm{16}\Rightarrow \\ $$$${p}_{\mathrm{2}} =\mathrm{17}\Rightarrow{p}_{\mathrm{1}} =\mathrm{2\&}{a}_{\mathrm{1}} =\mathrm{1} \\ $$$${n}=\mathrm{34},{p}_{\mathrm{2}} =\mathrm{5}\Rightarrow{a}_{\mathrm{2}} =\mathrm{1},\Rightarrow\left({p}_{\mathrm{1}} −\mathrm{1}\right).{p}_{\mathrm{1}} ^{{a}_{\mathrm{1}} −\mathrm{1}} =\mathrm{4} \\ $$$${p}_{\mathrm{1}} =\mathrm{2},{a}_{\mathrm{1}} =\mathrm{3}\Rightarrow{n}=\mathrm{2}^{\mathrm{3}} .\mathrm{5}=\mathrm{40} \\ $$$${p}_{\mathrm{2}} =\mathrm{3}\Rightarrow{a}_{\mathrm{2}} =\mathrm{1}\Rightarrow{p}_{\mathrm{1}} =\mathrm{1\&}{a}_{\mathrm{1}} =\mathrm{4}\Rightarrow{n}=\mathrm{2}^{\mathrm{4}} .\mathrm{3}=\mathrm{48} \\ $$$${r}=\mathrm{3}\Rightarrow\:\:\left({p}_{\mathrm{1}} −\mathrm{1}\right)\left({p}_{\mathrm{2}} −\mathrm{1}\right)\left({p}_{\mathrm{3}} −\mathrm{1}\right)\leqslant\mathrm{16}\Rightarrow \\ $$$$\left({p}_{\mathrm{1}} −\mathrm{1}\right)\left({p}_{\mathrm{2}} −\mathrm{1}\right)<\mathrm{4}\Rightarrow{p}_{\mathrm{2}} \leqslant\mathrm{5}\Rightarrow{p}_{\mathrm{2}} \in\left\{\mathrm{3},\mathrm{5}\right\} \\ $$$${p}_{\mathrm{3}} \leqslant\mathrm{9}\Rightarrow{p}_{\mathrm{3}} \in\left\{\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{7}\right\},{p}_{\mathrm{3}} >{p}_{\mathrm{2}} \Rightarrow{p}_{\mathrm{3}} \in\left\{\mathrm{5},\mathrm{7}\right\}\Rightarrow{p}_{\mathrm{3}} =\mathrm{5\&}{p}_{\mathrm{2}} =\mathrm{3} \\ $$$$\Rightarrow\left({p}_{\mathrm{1}} −\mathrm{1}\right){p}_{\mathrm{1}} ^{{a}_{\mathrm{1}} −\mathrm{1}} \mathrm{2}.\mathrm{3}^{{a}_{\mathrm{2}} −\mathrm{1}} .\mathrm{4}.\mathrm{5}^{{a}_{\mathrm{3}} −\mathrm{1}} =\mathrm{16}\Rightarrow \\ $$$${a}_{\mathrm{2}} ={a}_{\mathrm{3}} =\mathrm{1\&}{p}_{\mathrm{1}} =\mathrm{2\&}{a}_{\mathrm{1}} =\mathrm{2}\Rightarrow{n}=\mathrm{4}.\mathrm{3}.\mathrm{5}=\mathrm{60} \\ $$$$ \\ $$$$ \\ $$
Commented by 175 last updated on 04/Jun/20
nice job

Leave a Reply

Your email address will not be published. Required fields are marked *