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Find-the-number-of-x-1-2016-x-N-which-making-the-expression-4x-6-x-3-5-is-divided-by-11-




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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