1-dy-dx-d-y-dx-2-x-2-73-mod-216-3-If-2-x-1-2-x-2-2-x-3-2-x-n-80-000-where-x-1-x-2-x-3-x-n-are-distinct-whole-number-find-the-value-of-n-

Question Number 113060 by bemath last updated on 12/Sep/20

$$\:\left(\mathrm{1}\right)\:\sqrt{\frac{\mathrm{dy}}{\mathrm{dx}}}\:=\:\frac{\mathrm{d}\left(\sqrt{\mathrm{y}}\right)}{\mathrm{dx}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{x}^{\mathrm{2}} \:\equiv\:\mathrm{73}\:\left(\mathrm{mod}\:\mathrm{216}\right) \\ $$$$\left(\mathrm{3}\right)\:\mathrm{If}\:\mathrm{2}^{\mathrm{x}_{\mathrm{1}} } +\mathrm{2}^{\mathrm{x}_{\mathrm{2}} } +\mathrm{2}^{\mathrm{x}_{\mathrm{3}} } +…+\mathrm{2}^{\mathrm{x}_{\mathrm{n}} } =\mathrm{80},\mathrm{000} \\ $$$$\:\mathrm{where}\:\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,\mathrm{x}_{\mathrm{3}} ,…,\mathrm{x}_{\mathrm{n}} \:\mathrm{are}\:\mathrm{distinct} \\ $$$$\mathrm{whole}\:\mathrm{number}\:.\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{n} \\ $$

Answered by john santu last updated on 11/Sep/20

solve (√(dy/dx)) = ((d((√y)))/dx) (sol)(√(dy/dx)) = (dy/(2(√y) dx )) squaring both sides (dy/dx) = (1/(4y)) ((dy/dx))^2 ⇒ p^2 = 4yp p(p−4y) = 0 { ((p=0⇒(dy/dx) =0, y = C)),((p=4y⇒(dy/dx) = 4y ⇒ln y = 4x+k , y=C_1 e^(4x) )) :} ((JS)/(a math farmer))

$$\:\:\:{solve}\:\sqrt{\frac{{dy}}{{dx}}}\:=\:\frac{{d}\left(\sqrt{{y}}\right)}{{dx}} \\ $$$$\left({sol}\right)\sqrt{\frac{{dy}}{{dx}}}\:=\:\frac{{dy}}{\mathrm{2}\sqrt{{y}}\:{dx}\:} \\ $$$${squaring}\:{both}\:{sides} \\ $$$$\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{4}{y}}\:\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \Rightarrow\:{p}^{\mathrm{2}} \:=\:\mathrm{4}{yp} \\ $$$$\:{p}\left({p}−\mathrm{4}{y}\right)\:=\:\mathrm{0}\:\begin{cases}{{p}=\mathrm{0}\Rightarrow\frac{{dy}}{{dx}}\:=\mathrm{0},\:{y}\:=\:{C}}\\{{p}=\mathrm{4}{y}\Rightarrow\frac{{dy}}{{dx}}\:=\:\mathrm{4}{y}\:\Rightarrow\mathrm{ln}\:{y}\:=\:\mathrm{4}{x}+{k}\:,\:{y}={C}_{\mathrm{1}} {e}^{\mathrm{4}{x}} }\end{cases}\: \\ $$$$\:\:\:\frac{{JS}}{{a}\:{math}\:{farmer}} \\ $$

Commented by bemath last updated on 11/Sep/20

$$\checkmark\mathrm{thank}\:\mathrm{you} \\ $$

Answered by bobhans last updated on 11/Sep/20

(2) x^2 ≡73 (mod 216) because 216 = 2^3 .3^3 , we first solve the congruences → { ((x^2 ≡ 73 (mod 2^3 ))),((x^2 ≡73 (mod 3^3 ))) :} (a) regarding x^2 ≡73≡1 (mod 8) we obtain x≡ ±1 , ±3 (mod 8) (b) regarding x^2 ≡73≡100 (mod 27) we obtain x≡±10 (mod 27) using the Chinese Remainder Theorem with result from (a) &(b) we conclude x ≡ ±17, ±37, ±71, ± 91 (mod 216)

$$\left(\mathrm{2}\right)\:\mathrm{x}^{\mathrm{2}} \equiv\mathrm{73}\:\left(\mathrm{mod}\:\mathrm{216}\right) \\ $$$$\mathrm{because}\:\mathrm{216}\:=\:\mathrm{2}^{\mathrm{3}} .\mathrm{3}^{\mathrm{3}} \:,\:\mathrm{we}\:\mathrm{first}\:\mathrm{solve}\:\mathrm{the} \\ $$$$\mathrm{congruences}\:\:\rightarrow\begin{cases}{\mathrm{x}^{\mathrm{2}} \equiv\:\mathrm{73}\:\left(\mathrm{mod}\:\mathrm{2}^{\mathrm{3}} \right)}\\{\mathrm{x}^{\mathrm{2}} \equiv\mathrm{73}\:\left(\mathrm{mod}\:\mathrm{3}^{\mathrm{3}} \right)}\end{cases} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{regarding}\:\mathrm{x}^{\mathrm{2}} \equiv\mathrm{73}\equiv\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{8}\right) \\ $$$$\:\mathrm{we}\:\mathrm{obtain}\:\mathrm{x}\equiv\:\pm\mathrm{1}\:,\:\pm\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{8}\right) \\ $$$$\left(\mathrm{b}\right)\:\mathrm{regarding}\:\mathrm{x}^{\mathrm{2}} \equiv\mathrm{73}\equiv\mathrm{100}\:\left(\mathrm{mod}\:\mathrm{27}\right) \\ $$$$\:\:\mathrm{we}\:\mathrm{obtain}\:\mathrm{x}\equiv\pm\mathrm{10}\:\left(\mathrm{mod}\:\mathrm{27}\right) \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{Chinese}\:\mathrm{Remainder}\:\mathrm{Theorem} \\ $$$$\mathrm{with}\:\mathrm{result}\:\mathrm{from}\:\left(\mathrm{a}\right)\:\&\left(\mathrm{b}\right)\:\mathrm{we}\:\mathrm{conclude}\: \\ $$$$\mathrm{x}\:\equiv\:\pm\mathrm{17},\:\pm\mathrm{37},\:\pm\mathrm{71},\:\pm\:\mathrm{91}\:\left(\mathrm{mod}\:\mathrm{216}\right) \\ $$

Answered by bobhans last updated on 11/Sep/20

(3) consider 80,000_(10) =10,011,100,010,000,000_2 that say 80000=2^(16) +2^(13) +2^(12) +2^(11) +2^7 so n = 5

$$\left(\mathrm{3}\right)\:\mathrm{consider}\:\mathrm{80},\mathrm{000}_{\mathrm{10}} =\mathrm{10},\mathrm{011},\mathrm{100},\mathrm{010},\mathrm{000},\mathrm{000}_{\mathrm{2}} \\ $$$$\mathrm{that}\:\mathrm{say}\:\mathrm{80000}=\mathrm{2}^{\mathrm{16}} +\mathrm{2}^{\mathrm{13}} +\mathrm{2}^{\mathrm{12}} +\mathrm{2}^{\mathrm{11}} +\mathrm{2}^{\mathrm{7}} \\ $$$$\mathrm{so}\:\mathrm{n}\:=\:\mathrm{5} \\ $$

Commented by Rasheed.Sindhi last updated on 11/Sep/20