(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


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

$(1) \sqrt{\frac{dy}{dx}} = \frac{d (\sqrt{y})}{dx}$ $(2) x^{2} \equiv 73 (\mod 216)$ $(3) If 2^{x_{1}} + 2^{x_{2}} + 2^{x_{3}} + . . . + 2^{x_{n}} = 80, 000$ $where x_{1}, x_{2}, x_{3}, . . ., x_{n} are distinct$ $whole number . find the value of n$
	Answered by john santu last updated on 11/Sep/20
	$solve (√(dy/dx)) = ((d((√y)))/dx) (sol)(√(dy/dx)) = (dy/(2(√y) dx )) squaring both sides (dy/dx) = (1/(4y)) ((dy/dx))^2 ⇒ p^2 = 4yp p(p−4y) = 0 { ((p=0⇒(dy/dx) =0, y = C)),((p=4y⇒(dy/dx) = 4y ⇒ln y = 4x+k , y=C_1 e^(4x) )) :} ((JS)/(a math farmer))$
	$s o l v e \sqrt{\frac{d y}{d x}} = \frac{d (\sqrt{y})}{d x}$ $(s o l) \sqrt{\frac{d y}{d x}} = \frac{d y}{2 \sqrt{y} d x}$ $s q u a r i n g b o t h s i d e s$ $\frac{d y}{d x} = \frac{1}{4 y} {(\frac{d y}{d x})}^{2} \Rightarrow p^{2} = 4 y p$ $p (p - 4 y) = 0 {\begin{cases} p = 0 \Rightarrow \frac{d y}{d x} = 0, y = C \\ p = 4 y \Rightarrow \frac{d y}{d x} = 4 y \Rightarrow \ln y = 4 x + k, y = C_{1} e^{4 x} \end{cases}$ $\frac{J S}{a m a t h f a r m e r}$
	Commented by bemath last updated on 11/Sep/20

	$✓ thank you$
	Answered by bobhans last updated on 11/Sep/20
	$(2) x^2 ≡73 (mod 216) because 216 = 2^3 .3^3 , we first solve the congruences → { ((x^2 ≡ 73 (mod 2^3 ))),((x^2 ≡73 (mod 3^3 ))) :} (a) regarding x^2 ≡73≡1 (mod 8) we obtain x≡ ±1 , ±3 (mod 8) (b) regarding x^2 ≡73≡100 (mod 27) we obtain x≡±10 (mod 27) using the Chinese Remainder Theorem with result from (a) &(b) we conclude x ≡ ±17, ±37, ±71, ± 91 (mod 216)$
	$(2) x^{2} \equiv 73 (\mod 216)$ $because 216 = 2^{3} . 3^{3}, we first solve the$ $congruences \to {\begin{cases} x^{2} \equiv 73 (\mod 2^{3}) \\ x^{2} \equiv 73 (\mod 3^{3}) \end{cases}$ $(a) regarding x^{2} \equiv 73 \equiv 1 (\mod 8)$ $we obtain x \equiv \pm 1, \pm 3 (\mod 8)$ $(b) regarding x^{2} \equiv 73 \equiv 100 (\mod 27)$ $we obtain x \equiv \pm 10 (\mod 27)$ $using the Chinese Remainder Theorem$ $with result from (a) & (b) we conclude$ $x \equiv \pm 17, \pm 37, \pm 71, \pm 91 (\mod 216)$
	Answered by bobhans last updated on 11/Sep/20

	$(3) consider 80, 000_{10} = 10, 011, 100, 010, 000, 000_{2}$ $that say 80000 = 2^{16} + 2^{13} + 2^{12} + 2^{11} + 2^{7}$ $so n = 5$
	Commented by Rasheed.Sindhi last updated on 11/Sep/20

	$e^{X} c e l l e n t!$
	Commented by 1549442205PVT last updated on 12/Sep/20

	$Number in the question is 8000 ?$
	Commented by bemath last updated on 12/Sep/20

	$80, 000 sir$
	Commented by Rasheed.Sindhi last updated on 12/Sep/20

	$B u t t h e n u m b e r i s 8000 .$
	Commented by bemath last updated on 12/Sep/20

	$sorry sir typo$
	Commented by bemath last updated on 12/Sep/20

	$wkwkwkek . . =====$
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