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Question Number 198785 by sonukgindia last updated on 24/Oct/23

	Answered by deleteduser1 last updated on 24/Oct/23

	$\frac{1}{197} \equiv x (m o d 3000) \Rightarrow 197 x \equiv 1 (m o d 3000)$ $197 x \equiv 1 (m o d 3) \Rightarrow - x \overset{3}{\equiv} 1 \Rightarrow x \overset{3}{\equiv} 2$ $197 x \equiv 1 (m o d 8) \Rightarrow 5 x \equiv 1 (m o d 8) \Rightarrow x \overset{8}{\equiv} 5$ $197 x \equiv 1 (m o d 125) \Rightarrow 72 x \equiv 1 (m o d 125)$ $\Rightarrow - 53 x \equiv 1 (m o d 125) \Rightarrow [3 x \overset{25}{\equiv} - 1 \Rightarrow x \overset{25}{\equiv} 8]$ $53 x \overset{125}{\equiv} 124 + 125 \times 13 \Rightarrow x \equiv 33 (m o d 125)$ $x = 3 k + 2; x = 8 q + 5 \Rightarrow 8 q + 5 \equiv 2 (m o d 3)$ $\Rightarrow q \equiv 0 (m o d 3) \Rightarrow x = 24 v + 5 \equiv 33 (m o d 125)$ $\Rightarrow 24 v \equiv 28 (m o d 125) \Rightarrow 6 v \overset{125}{\equiv} 7 \Rightarrow v \equiv 22 (m o d 125)$ $\Rightarrow x = 24 (125 y + 22) + 5 \Rightarrow x \equiv 24 \times 22 + 5 = 533$ $\Rightarrow x \equiv 533 (m o d 3000)$
	Commented by qaz last updated on 24/Oct/23

	$197 x \equiv 1 (m o d 3000)$ $\Rightarrow 197 x + 3000 a = 1 \Rightarrow 3000 a = 45 a \equiv 1 (m o d 197)$ $\Rightarrow 45 a + 197 b = 1 \Rightarrow 197 b = 17 b \equiv 1 (m o d 45)$ $\Rightarrow 17 b + 45 c = 1 \Rightarrow 45 c = 11 c \equiv 1 (m o d 17)$ $\Rightarrow 11 c + 17 d = 1 \Rightarrow 17 d \equiv 6 d \equiv 1 (m o d 11)$ $\Rightarrow d \equiv 2 (m o d 11)$ $\Rightarrow c \equiv - 3 (m o d 17) \Rightarrow b \equiv 8 (m o d 45) \Rightarrow a \equiv - 35 (m o d 197) \Rightarrow x \equiv 533 (m o d 3000)$
	Answered by deleteduser1 last updated on 24/Oct/23

	$2 . n^{5} \equiv 155^{777^{888}} \equiv {(- 9)}^{777^{888}} \overset{41}{\equiv} - {(9)}^{777^{888}}$ $777^{888} \equiv {(17)}^{888} (m o d 40)$ $17^{888} = 17^{16 (55) + 8} \equiv 7^{8} (m o d 40) \equiv 9^{4} = 81 = 1 (m o d 40)$ $\Rightarrow - {(9)}^{777^{888}} = - {(9)}^{40 q + 1} \equiv - {(9)}^{40 q} 8 \equiv - 9 (m o d 41)$ $\Rightarrow n^{5} \equiv 32 (m o d 41) \Rightarrow n \equiv 2 (m o d 41)$
	Answered by deleteduser1 last updated on 24/Oct/23
	$3. n^5 ≡6^k (mod 41)⇒6^(5q) ≡6^k (mod 41) 6^(5q−k) ≡6^(40) (mod 41)⇒5q≡k(mod 40) ⇒k≡5q(mod 40)⇒k≡5q(mod 41) where 5q≤40 ⇒x≡6^0 ,6^5 ,6^(10) ,6^(15) ,6^(20) ,6^(25) ,6^(30) ,6^(35) x∈{[1],[3],[9],[14],[27],[32],[38]}$
	$3 . n^{5} \equiv 6^{k} (m o d 41) \Rightarrow 6^{5 q} \equiv 6^{k} (m o d 41)$ $6^{5 q - k} \equiv 6^{40} (m o d 41) \Rightarrow 5 q \equiv k (m o d 40)$ $\Rightarrow k \equiv 5 q (m o d 40) \Rightarrow k \equiv 5 q (m o d 41) w h e r e 5 q ⩽ 40$ $\Rightarrow x \equiv 6^{0}, 6^{5}, 6^{10}, 6^{15}, 6^{20}, 6^{25}, 6^{30}, 6^{35}$ $x \in {[1], [3], [9], [14], [27], [32], [38]}$
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