Find the number of x ∈ [1, 2016 ] , x ∈ N which making the expression 4x^6 + x^3 + 5 is divided by 11 .


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Question Number 157599 by naka3546 last updated on 25/Oct/21

$F i n d t h e n u m b e r o f x \in [1, 2016], x \in N$ $w h i c h m a k i n g t h e e x p r e s s i o n 4 x^{6} + x^{3} + 5 i s d i v i d e d b y 11 .$
	Answered by TheSupreme last updated on 25/Oct/21

	$4 x^{6} + x^{3} + 5 \equiv 0 (11)$ $4 x^{6} + x^{3} \equiv 6 (11)$ $x^{3} (4 x^{3} + 1) \equiv 6 (11)$ $x^{3} = A$ $A (4 A + 1) \equiv 6 (11)$ $A \equiv 1 (11) 5$ $A \equiv 2 (11) 7$ $A \equiv 3 (11) 6$ $A \equiv 4 (11) 2$ $A \equiv 5 (11) 8$ $A \equiv 6 (11) 7$ $A \equiv 7 (11) 5$ $A \equiv 8 (11) 0$ $A \equiv 9 (11) 3$ $A \equiv 1 (11) 0 3$ $A = 0 (11) 0$ $. . .$ $A \equiv 3 (11)$ $x^{3} \equiv 3 (11)$ $x \equiv 1 (11) 1$ $x \equiv 2 (11) 4$ $x \equiv 3 (11) 9$ $x \equiv 4 (11) 5$ $x \equiv 5 (11) 3$ $x \equiv 6 (11) 3$ $x \equiv 7 (11) 5$ $x \equiv 8 (11) 9$ $x \equiv 9 (11) 4$ $x \equiv 10 (11) 1$ $x \equiv 0 (11) 0$ $x \equiv 5 (11) a n d x \equiv 6 (11)$ $D (2016, 5) = ⌊ \frac{2016 - 5}{11} ⌋ + ⌊ \frac{2016 - 6}{11} ⌋ = 182 + 182 = 364$
	Commented by naka3546 last updated on 25/Oct/21

	$T h a n k y o u s o m u c h, s i r .$
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