Question-217897

Question Number 217897 by Unhombre last updated on 23/Mar/25

Commented by Hanuda354 last updated on 23/Mar/25

A = ?

Commented by Unhombre last updated on 23/Mar/25

A = 0 / 5243 a n d a s t r a i g h t l i n e a b o v e i t

m e a n i n g 0 / 52435243 \dots

Commented by mr W last updated on 23/Mar/25

B_{1} = 5343

B_{2} = 52435243

\dots

B_{n} = \underset{n t i m e s 5343}{52435243 \dots 5243}

\lim_{n \to \infty} B_{n} = \infty

A = \frac{0}{\lim_{n \to \infty} B_{n}} = 0

\Rightarrow \frac{9}{10 A - 5} = - \frac{9}{5}

Commented by mr W last updated on 23/Mar/25

y o u c o u l d a l s o g i v e A = 1 / \overset{―}{5243}, o r

A = 12345 / \overset{―}{5243}, e t c .

Commented by Unhombre last updated on 23/Mar/25

t h a n k s b u t t h e o p t i o n s a r e :

27

37

38

39

Commented by mr W last updated on 23/Mar/25

t h e n t h e q u e s t i o n i s w r o n g o r t h e

a n s w e r g i v e n i s w r o n g .

i f i t i s m e a n t A = 0 . \overset{―}{5243}, t h e n

10000 A = 5243 + A

A = \frac{5243}{9999}

\frac{9}{10 A - 5} = \frac{9}{10 \times \frac{5243}{9999} - 5} \approx 37

Commented by Unhombre last updated on 23/Mar/25

y e a h t h e a n s w e r i s e x a c t l y 37

t h a n k s

Answered by mahdipoor last updated on 23/Mar/25

A = 0.5243 (1 + 10^{- 4} + 10^{- 8} + 10^{- 12} + \dots) \times \frac{1 - 10^{- 4}}{1 - 10^{- 4}}

0.5243 \frac{1 - 10^{- \infty}}{1 - 10^{- 4}} (= \frac{5243}{9999} =)

\frac{9}{10 A - 5} = \frac{89991}{2435} \approx 36.96