2^a +4^b +8^c =328 find a,b and c when(a,b,c)is natual number


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Question Number 191663 by mathlove last updated on 28/Apr/23

$2^{a} + 4^{b} + 8^{c} = 328$ $f i n d a, b a n d c$ $w h e n (a, b, c) i s n a t u a l n u m b e r$
	Answered by Rasheed.Sindhi last updated on 28/Apr/23
	$2^a +4^b +8^c =328 2^a +2^(2b) +2^(3c) =328=101001000_2 =2^8 +2^6 +2^3 3c=3,6⇒c=1,2 2b=6,8⇒b=3,4 a=3,6,8 c= { ((1⇒b= { ((3⇒a=8)),((4⇒a=6)) :})),((2⇒b=4⇒a=3)) :} (a,b,c)=(3,4,2),(6,4,1),(8,3,1)$
	$2^{a} + 4^{b} + 8^{c} = 328$ $2^{a} + 2^{2 b} + 2^{3 c} = 328 = 101001000_{2}$ $= 2^{8} + 2^{6} + 2^{3}$ $3 c = 3, 6 \Rightarrow c = 1, 2$ $2 b = 6, 8 \Rightarrow b = 3, 4$ $a = 3, 6, 8$ $c = {\begin{cases} 1 \Rightarrow b = {\begin{cases} 3 \Rightarrow a = 8 \\ 4 \Rightarrow a = 6 \end{cases} \\ 2 \Rightarrow b = 4 \Rightarrow a = 3 \end{cases}$ $(a, b, c) = (3, 4, 2), (6, 4, 1), (8, 3, 1)$
	Commented by BaliramKumar last updated on 28/Apr/23

	$Nice solution Sir$
	Commented by mathlove last updated on 28/Apr/23

	$t h a n k s s i r$
	Answered by BaliramKumar last updated on 28/Apr/23

	$2^{a} + 2^{2 b} + 2^{3 c} = 2^{3} \times 41$ $2^{3 c} (2^{a - 3 c} + 2^{2 b - 3 c} + 1) = 2^{3} \times 41$ $c = 1$ $2^{a - 3} + 2^{2 b - 3} + 1 = 41$ $2^{a - 3} + 2^{2 b - 3} = 40$ $2^{a - 3} + 2^{2 b - 3} = 2^{3} \times 5$ $2^{a} + 2^{2 b} = 2^{6} \times 5$ $2^{2 b} (2^{a - 2 b} + 1) = 2^{6} \times 5$ $b = 3$ $2^{a - 2 \times 3} + 1 = 5$ $2^{a - 6} = 5 - 1 = 4 = 2^{2}$ $a - 6 = 2$ $a = 8$
	Commented by mathlove last updated on 28/Apr/23

	$t h a n k s$
	Answered by mehdee42 last updated on 28/Apr/23

	$n u m b e r 328 i s m a d e b y c o m b i n i n g t h e s u m o f t h r e e n u m b e r s 256, 64 & 8 . a c c o r d i n g t o t h e e x p e r s s i o n s i n t h e g i v e n e q u t i o n, w e h s v e t h e f o l l o w i n g t h r e e s t a t e s .$ $(2^{a}, 4^{b}, 8^{c}) = (256, 64, 8) = (64, 256, 8) = (8, 256, 64)$ $\Rightarrow (a, b, c) = (8, 3, 1) \lor (6, 4, 1) \lor (3, 4, 2)$
	Answered by LOSER last updated on 29/Apr/23

	$2^{a}; 4^{b}; 8^{c} > 0 \forall a; b; c \in R$ $\Rightarrow 8^{c} < 328 \Rightarrow c = 2; c = 1; c = 0$ $∙ c = 2$ $\Rightarrow 2^{a} + 4^{b} = 264 \Rightarrow 4^{b} < 264$ $\Rightarrow b = 4; b = 3; b = 2; b = 1; b = 0$ $* b = 4 \Rightarrow a = 3$ $* b = 3; b = 2; b = 1; b = 0 \Rightarrow ∄ a$ $∙ c = 1$ $\Rightarrow 2^{a} + 4^{b} = 320 \Rightarrow 4^{b} < 320$ $\Rightarrow b = 4; b = 3; b = 2; b = 1; b = 0$ $* b = 4 \Rightarrow a = 6$ $* b = 3 \Rightarrow a = 8$ $* b = 2; b = 1; b = 0 \Rightarrow ∄ a$ $∙ c = 0$ $\Rightarrow 2^{a} + 4^{b} = 327$ $\Rightarrow b = 4; b = 3; b = 2; b = 1; b = 0 \Rightarrow ∄ a$ $\Rightarrow (a; b; c) = (3; 4; 2); (6; 4; 1); (8; 3; 1)$
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