2-a-4-b-8-c-328-find-a-b-and-c-when-a-b-c-is-natual-number-

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^{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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