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Question Number 199381 by Mingma last updated on 02/Nov/23

	Answered by witcher3 last updated on 02/Nov/23
	$(202)=2.101 (((201)!)/k)≡0[202]0,∀k∈{1,......201)−{2,101) J≡((201!)/(101))+((201!)/2)[202] ((201!)/2)=202.3.2.Π_(k=5,k≠101) ^(201) k≡0[202] ((201!)/(101))=100!.Π_1 ^(100) (101+k) wilson theorm⇒101 is prime⇒(101−1)!≡−1(101) Π_1 ^(100) (101+k)≡Π(k)[101] ⇔(101−1)!≡−1[101] ⇒((201!)/(101))≡−1.−1[101]≡1[101] and ((201!)/(101))≡0[2] x≡0[2] x≡1[101] x=2k 2k≡1[101]⇒k≡51[101] x=202k+102 J=202n+102$
	$(202) = 2.101$ $\frac{(201)!}{k} \equiv 0 [202] 0, \forall k \in {1, . . . . . . 201) - {2, 101)$ $J \equiv \frac{201!}{101} + \frac{201!}{2} [202]$ $\frac{201!}{2} = 202.3.2 . \underset{k = 5, k \neq 101}{\prod^{201}} k \equiv 0 [202]$ $\frac{201!}{101} = 100! . \underset{1}{\prod^{100}} (101 + k)$ $wilson theorm \Rightarrow 101 is prime \Rightarrow (101 - 1)! \equiv - 1 (101)$ $\underset{1}{\prod^{100}} (101 + k) \equiv Π (k) [101]$ $\Leftrightarrow (101 - 1)! \equiv - 1 [101]$ $\Rightarrow \frac{201!}{101} \equiv - 1 . - 1 [101] \equiv 1 [101]$ $and \frac{201!}{101} \equiv 0 [2]$ $x \equiv 0 [2]$ $x \equiv 1 [101]$ $x = 2 k$ $2 k \equiv 1 [101] \Rightarrow k \equiv 51 [101]$ $x = 202 k + 102$ $J = 202 n + 102$
	Commented by Mingma last updated on 03/Nov/23
	Perfect ��
	Commented by witcher3 last updated on 04/Nov/23

	$thank You$
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