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Question Number 96671 by 175 last updated on 03/Jun/20

	Answered by maths mind last updated on 03/Jun/20
	$n=Π_(i=1) ^r p_i ^a_i ⇒ ϕ(n)=Π_(i=1) ^r (p_i −1)p_i ^(a_i −1) =16=2^4 p_1 <p_2 <p_3 .....<p_r ⇒r≤3& p_r ≤17 r=1⇒ϕ(n)=(p_1 −1)p_1 ^(a_1 −1) =16⇒p_1 =2 ,a_1 =5⇔n=32 r=2⇒(p_1 −1)p_1 ^(a_1 −1) .(p_2 −1).p_2 ^(a_2 −1) =16⇒ p_2 =17⇒p_1 =2&a_1 =1 n=34,p_2 =5⇒a_2 =1,⇒(p_1 −1).p_1 ^(a_1 −1) =4 p_1 =2,a_1 =3⇒n=2^3 .5=40 p_2 =3⇒a_2 =1⇒p_1 =1&a_1 =4⇒n=2^4 .3=48 r=3⇒ (p_1 −1)(p_2 −1)(p_3 −1)≤16⇒ (p_1 −1)(p_2 −1)<4⇒p_2 ≤5⇒p_2 ∈{3,5} p_3 ≤9⇒p_3 ∈{2,3,5,7},p_3 >p_2 ⇒p_3 ∈{5,7}⇒p_3 =5&p_2 =3 ⇒(p_1 −1)p_1 ^(a_1 −1) 2.3^(a_2 −1) .4.5^(a_3 −1) =16⇒ a_2 =a_3 =1&p_1 =2&a_1 =2⇒n=4.3.5=60$
	$n = \underset{i = 1}{\prod^{r}} p_{i}^{a_{i}} \Rightarrow$ $φ (n) = \underset{i = 1}{\prod^{r}} (p_{i} - 1) p_{i}^{a_{i} - 1} = 16 = 2^{4}$ $p_{1} < p_{2} < p_{3} . . . . . < p_{r} \Rightarrow r ⩽ 3 & p_{r} ⩽ 17$ $r = 1 \Rightarrow φ (n) = (p_{1} - 1) p_{1}^{a_{1} - 1} = 16 \Rightarrow p_{1} = 2, a_{1} = 5 \Leftrightarrow n = 32$ $r = 2 \Rightarrow (p_{1} - 1) p_{1}^{a_{1} - 1} . (p_{2} - 1) . p_{2}^{a_{2} - 1} = 16 \Rightarrow$ $p_{2} = 17 \Rightarrow p_{1} = 2 & a_{1} = 1$ $n = 34, p_{2} = 5 \Rightarrow a_{2} = 1, \Rightarrow (p_{1} - 1) . p_{1}^{a_{1} - 1} = 4$ $p_{1} = 2, a_{1} = 3 \Rightarrow n = 2^{3} . 5 = 40$ $p_{2} = 3 \Rightarrow a_{2} = 1 \Rightarrow p_{1} = 1 & a_{1} = 4 \Rightarrow n = 2^{4} . 3 = 48$ $r = 3 \Rightarrow (p_{1} - 1) (p_{2} - 1) (p_{3} - 1) ⩽ 16 \Rightarrow$ $(p_{1} - 1) (p_{2} - 1) < 4 \Rightarrow p_{2} ⩽ 5 \Rightarrow p_{2} \in {3, 5}$ $p_{3} ⩽ 9 \Rightarrow p_{3} \in {2, 3, 5, 7}, p_{3} > p_{2} \Rightarrow p_{3} \in {5, 7} \Rightarrow p_{3} = 5 & p_{2} = 3$ $\Rightarrow (p_{1} - 1) p_{1}^{a_{1} - 1} 2 . 3^{a_{2} - 1} . 4 . 5^{a_{3} - 1} = 16 \Rightarrow$ $a_{2} = a_{3} = 1 & p_{1} = 2 & a_{1} = 2 \Rightarrow n = 4.3.5 = 60$
	Commented by 175 last updated on 04/Jun/20
	nice job
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