Find-a-b-c-N-2-a-4-b-8-c-328-

Question Number 182216 by Acem last updated on 05/Dec/22

F i n d a, b, c \in N; 2^{a} + 4^{b} + 8^{c} = 328

Commented by SANOGO last updated on 06/Dec/22

t h a n k y o u

Commented by Acem last updated on 06/Dec/22

{Y o u}^{'} r e v e r y m u c h w e l c o m e M r . S a n o g o

y o u c a n s e e t h e w a y t o f i n d t h e p o w e r s b e l o w

e s p e c i a l y i f t h e 2 n d s i d e w a s b i g g e r t h a n 328

Answered by HeferH last updated on 06/Dec/22

R e m i n d e d m e o f t h e b i n a r y s y s t e m :

328 = 2^{8} + 2^{6} + 2^{3} = 256 + 64 + 8

2^{a} + 2^{2 b} + 2^{3 c} = 2^{8} + 2^{6} + 2^{3}

O n e s o l u t i o n w o u l d b e :

a = 8, b = 3, c = 1;

A n o t h e r o n e :

a = 3, b = 4, c = 2;

a n d a n o t h e r o n e :

a = 6, b = 8, c = 1

\begin{array}{ccccccc} a & 2 b & 3 c \\ 8 & 6 & 3 \\ 3 & 6 & 8 \\ 6 & 3 & 8 \\ 8 & 3 & 3 \\ 6 & 8 & 3 \\ 3 & 8 & 6 \end{array}

S i n c e a, b, c \in N

Commented by Acem last updated on 06/Dec/22

T h a n k s f r i e n d! v e r y w e l l, a n d p l e a s e r e c h e c k,

c a u s e {t h e r e}^{'} s s t i l l a n o t h e r s o l u t i o n, h o p e y o u

c a t c h i t!

Commented by Acem last updated on 06/Dec/22

N o t e : w h a t i f t h e s e c o n d s i d e w a s b i g g e r t h a n 328

I t h i n k t h a t w i l l b e d i f f i c u l t t o s e g m e n t i t d i r c e t l y

a s y o u h a v e d o n e a b o v e . S o m e t h i n g m u s t b e d o n e

t o f i n d t h e p o w e r s .

Commented by HeferH last updated on 06/Dec/22

I f o r g o r r

a = 6, b = 4 a n d c = 1

Commented by Acem last updated on 06/Dec/22

Y e s S i r! T h a n k s s s

Commented by Acem last updated on 06/Dec/22

H i S i r! H e r e i t i s t h e w a y t o f i n d t h e p o w e r s, {i t}^{'} s

b e l o w . T h a n k y o u a g a i n!

Answered by Rasheed.Sindhi last updated on 06/Dec/22

2^{a} + 2^{2 b} + 2^{3 c} = 256 + 64 + 8

= 2^{8} + 2^{6} + 2^{3}

Possible values :

a = \overset{✓}{8}, \overset{✓}{6}, \overset{✓}{3}; 2 \underset{b = 4, 3}{b = 8, 6}, \overset{\times}{3}; 3 \underset{c = 2, 1}{c = 8, 6}, \overset{✓}{3}

a = {\begin{cases} 8 \Rightarrow b = 3 \Rightarrow c = 1 \\ 6 \Rightarrow b = 4 \Rightarrow c = 1 \\ 3 \Rightarrow b = {\begin{cases} 4 \Rightarrow c = 2 \\ 3 \Rightarrow 3 c \neq 8 \times \end{cases} \end{cases}

(a, b, c) = (8, 3, 1), (6, 4, 1), (3, 4, 2)

Commented by Acem last updated on 06/Dec/22

T h a n k s S i r!

Answered by Acem last updated on 06/Dec/22

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