Question Number 199381 by Mingma last updated on 02/Nov/23
Answered by witcher3 last updated on 02/Nov/23
![(202)=2.101 (((201)!)/k)≡0[202]0,∀k∈{1,......201)−{2,101) J≡((201!)/(101))+((201!)/2)[202] ((201!)/2)=202.3.2.Π_(k=5,k≠101) ^(201) k≡0[202] ((201!)/(101))=100!.Π_1 ^(100) (101+k) wilson theorm⇒101 is prime⇒(101−1)!≡−1(101) Π_1 ^(100) (101+k)≡Π(k)[101] ⇔(101−1)!≡−1[101] ⇒((201!)/(101))≡−1.−1[101]≡1[101] and ((201!)/(101))≡0[2] x≡0[2] x≡1[101] x=2k 2k≡1[101]⇒k≡51[101] x=202k+102 J=202n+102](https://www.tinkutara.com/question/Q199383.png)
$$\left(\mathrm{202}\right)=\mathrm{2}.\mathrm{101} \\ $$$$\frac{\left(\mathrm{201}\right)!}{\mathrm{k}}\equiv\mathrm{0}\left[\mathrm{202}\right]\mathrm{0},\forall\mathrm{k}\in\left\{\mathrm{1},……\mathrm{201}\right)−\left\{\mathrm{2},\mathrm{101}\right) \\ $$$$\mathrm{J}\equiv\frac{\mathrm{201}!}{\mathrm{101}}+\frac{\mathrm{201}!}{\mathrm{2}}\left[\mathrm{202}\right] \\ $$$$\frac{\mathrm{201}!}{\mathrm{2}}=\mathrm{202}.\mathrm{3}.\mathrm{2}.\underset{\mathrm{k}=\mathrm{5},\mathrm{k}\neq\mathrm{101}} {\overset{\mathrm{201}} {\prod}}\mathrm{k}\equiv\mathrm{0}\left[\mathrm{202}\right] \\ $$$$\frac{\mathrm{201}!}{\mathrm{101}}=\mathrm{100}!.\underset{\mathrm{1}} {\overset{\mathrm{100}} {\prod}}\left(\mathrm{101}+\mathrm{k}\right) \\ $$$$\mathrm{wilson}\:\mathrm{theorm}\Rightarrow\mathrm{101}\:\mathrm{is}\:\mathrm{prime}\Rightarrow\left(\mathrm{101}−\mathrm{1}\right)!\equiv−\mathrm{1}\left(\mathrm{101}\right) \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{100}} {\prod}}\left(\mathrm{101}+\mathrm{k}\right)\equiv\Pi\left(\mathrm{k}\right)\left[\mathrm{101}\right] \\ $$$$\Leftrightarrow\left(\mathrm{101}−\mathrm{1}\right)!\equiv−\mathrm{1}\left[\mathrm{101}\right] \\ $$$$\Rightarrow\frac{\mathrm{201}!}{\mathrm{101}}\equiv−\mathrm{1}.−\mathrm{1}\left[\mathrm{101}\right]\equiv\mathrm{1}\left[\mathrm{101}\right] \\ $$$$\mathrm{and}\:\frac{\mathrm{201}!}{\mathrm{101}}\equiv\mathrm{0}\left[\mathrm{2}\right] \\ $$$$\mathrm{x}\equiv\mathrm{0}\left[\mathrm{2}\right] \\ $$$$\mathrm{x}\equiv\mathrm{1}\left[\mathrm{101}\right] \\ $$$$\mathrm{x}=\mathrm{2k} \\ $$$$\mathrm{2k}\equiv\mathrm{1}\left[\mathrm{101}\right]\Rightarrow\mathrm{k}\equiv\mathrm{51}\left[\mathrm{101}\right] \\ $$$$\mathrm{x}=\mathrm{202k}+\mathrm{102} \\ $$$$\mathrm{J}=\mathrm{202n}+\mathrm{102} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Mingma last updated on 03/Nov/23
Perfect ��
Commented by witcher3 last updated on 04/Nov/23
$$\mathrm{thank}\:\mathrm{You} \\ $$$$ \\ $$