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

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

$${Find}\:\:{the}\:\:{number}\:\:{of}\:\:{x}\:\in\:\left[\mathrm{1},\:\mathrm{2016}\:\right]\:\:,\:\:{x}\:\in\:\mathbb{N} \\ $$$${which}\:\:{making}\:\:{the}\:\:{expression}\:\:\mathrm{4}{x}^{\mathrm{6}} \:+\:\:{x}^{\mathrm{3}} \:+\:\mathrm{5}\:\:\:{is}\:\:{divided}\:\:\:{by}\:\:\mathrm{11}\:. \\ $$

Answered by TheSupreme last updated on 25/Oct/21

4x^6 +x^3 +5≡0(11) 4x^6 +x^3 ≡6(11) x^3 (4x^3 +1)≡6(11) x^3 =A A(4A+1)≡6(11) A≡1(11) 5 A≡2(11) 7 A≡3(11) 6 A≡4(11) 2 A≡5(11) 8 A≡6(11) 7 A≡7(11) 5 A≡8(11) 0 A≡9(11) 3 A≡1(11)0 3 A=0(11) 0 ... A≡3(11) x^3 ≡3(11) x≡1(11) 1 x≡2(11) 4 x≡3(11) 9 x≡4(11) 5 x≡5(11) 3 x≡6(11) 3 x≡7(11) 5 x≡8(11) 9 x≡9(11) 4 x≡10(11) 1 x≡0(11) 0 x≡5(11) and x≡6(11) D(2016,5)=⌊((2016−5)/(11))⌋+⌊((2016−6)/(11))⌋=182+182=364

$$\mathrm{4}{x}^{\mathrm{6}} +{x}^{\mathrm{3}} +\mathrm{5}\equiv\mathrm{0}\left(\mathrm{11}\right) \\ $$$$\mathrm{4}{x}^{\mathrm{6}} +{x}^{\mathrm{3}} \equiv\mathrm{6}\left(\mathrm{11}\right) \\ $$$${x}^{\mathrm{3}} \left(\mathrm{4}{x}^{\mathrm{3}} +\mathrm{1}\right)\equiv\mathrm{6}\left(\mathrm{11}\right) \\ $$$${x}^{\mathrm{3}} ={A} \\ $$$${A}\left(\mathrm{4}{A}+\mathrm{1}\right)\equiv\mathrm{6}\left(\mathrm{11}\right) \\ $$$$ \\ $$$${A}\equiv\mathrm{1}\left(\mathrm{11}\right)\:\mathrm{5} \\ $$$${A}\equiv\mathrm{2}\left(\mathrm{11}\right)\:\mathrm{7} \\ $$$${A}\equiv\mathrm{3}\left(\mathrm{11}\right)\:\mathrm{6} \\ $$$${A}\equiv\mathrm{4}\left(\mathrm{11}\right)\:\mathrm{2} \\ $$$${A}\equiv\mathrm{5}\left(\mathrm{11}\right)\:\mathrm{8} \\ $$$${A}\equiv\mathrm{6}\left(\mathrm{11}\right)\:\mathrm{7} \\ $$$${A}\equiv\mathrm{7}\left(\mathrm{11}\right)\:\mathrm{5} \\ $$$${A}\equiv\mathrm{8}\left(\mathrm{11}\right)\:\mathrm{0} \\ $$$${A}\equiv\mathrm{9}\left(\mathrm{11}\right)\:\mathrm{3} \\ $$$${A}\equiv\mathrm{1}\left(\mathrm{11}\right)\mathrm{0}\:\mathrm{3} \\ $$$${A}=\mathrm{0}\left(\mathrm{11}\right)\:\mathrm{0} \\ $$$$... \\ $$$${A}\equiv\mathrm{3}\left(\mathrm{11}\right) \\ $$$$ \\ $$$${x}^{\mathrm{3}} \equiv\mathrm{3}\left(\mathrm{11}\right) \\ $$$$ \\ $$$${x}\equiv\mathrm{1}\left(\mathrm{11}\right)\:\mathrm{1} \\ $$$${x}\equiv\mathrm{2}\left(\mathrm{11}\right)\:\mathrm{4} \\ $$$${x}\equiv\mathrm{3}\left(\mathrm{11}\right)\:\mathrm{9} \\ $$$${x}\equiv\mathrm{4}\left(\mathrm{11}\right)\:\mathrm{5} \\ $$$${x}\equiv\mathrm{5}\left(\mathrm{11}\right)\:\mathrm{3} \\ $$$${x}\equiv\mathrm{6}\left(\mathrm{11}\right)\:\mathrm{3} \\ $$$${x}\equiv\mathrm{7}\left(\mathrm{11}\right)\:\mathrm{5} \\ $$$${x}\equiv\mathrm{8}\left(\mathrm{11}\right)\:\mathrm{9} \\ $$$${x}\equiv\mathrm{9}\left(\mathrm{11}\right)\:\mathrm{4} \\ $$$${x}\equiv\mathrm{10}\left(\mathrm{11}\right)\:\mathrm{1} \\ $$$${x}\equiv\mathrm{0}\left(\mathrm{11}\right)\:\mathrm{0} \\ $$$$ \\ $$$${x}\equiv\mathrm{5}\left(\mathrm{11}\right)\:{and}\:{x}\equiv\mathrm{6}\left(\mathrm{11}\right) \\ $$$${D}\left(\mathrm{2016},\mathrm{5}\right)=\lfloor\frac{\mathrm{2016}−\mathrm{5}}{\mathrm{11}}\rfloor+\lfloor\frac{\mathrm{2016}−\mathrm{6}}{\mathrm{11}}\rfloor=\mathrm{182}+\mathrm{182}=\mathrm{364} \\ $$

Commented by naka3546 last updated on 25/Oct/21

Thank you so much, sir.

$${Thank}\:\:{you}\:\:{so}\:\:{much},\:\:{sir}. \\ $$

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