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
$${A}\:=\:\mathrm{0}/\mathrm{5243}\:{and}\:{a}\:{straight}\:{line}\:{above}\:{it} \\ $$$${meaning}\:\mathrm{0}/\mathrm{52435243}… \\ $$
Commented by mr W last updated on 23/Mar/25
$${B}_{\mathrm{1}} =\mathrm{5343} \\ $$$${B}_{\mathrm{2}} =\mathrm{52435243} \\ $$$$… \\ $$$${B}_{{n}} =\underset{{n}\:{times}\:\mathrm{5343}} {\mathrm{52435243}…\mathrm{5243}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{B}_{{n}} =\infty \\ $$$${A}=\frac{\mathrm{0}}{\underset{{n}\rightarrow\infty} {\mathrm{lim}}{B}_{{n}} }=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{9}}{\mathrm{10}{A}−\mathrm{5}}=−\frac{\mathrm{9}}{\mathrm{5}} \\ $$
Commented by mr W last updated on 23/Mar/25
$${you}\:{could}\:{also}\:{give}\:{A}=\mathrm{1}/\overline {\mathrm{5243}},\:{or} \\ $$$${A}=\mathrm{12345}/\overline {\mathrm{5243}},\:{etc}. \\ $$
Commented by Unhombre last updated on 23/Mar/25
$${thanks}\:{but}\:{the}\:{options}\:{are}: \\ $$$$\mathrm{27} \\ $$$$\mathrm{37} \\ $$$$\mathrm{38} \\ $$$$\mathrm{39} \\ $$
Commented by mr W last updated on 23/Mar/25
$${then}\:{the}\:{question}\:{is}\:{wrong}\:{or}\:{the} \\ $$$${answer}\:{given}\:{is}\:{wrong}. \\ $$$$ \\ $$$${if}\:{it}\:{is}\:{meant}\:{A}=\mathrm{0}.\overline {\mathrm{5243}},\:{then} \\ $$$$\mathrm{10000}{A}=\mathrm{5243}+{A} \\ $$$${A}=\frac{\mathrm{5243}}{\mathrm{9999}} \\ $$$$\frac{\mathrm{9}}{\mathrm{10}{A}−\mathrm{5}}=\frac{\mathrm{9}}{\mathrm{10}×\frac{\mathrm{5243}}{\mathrm{9999}}−\mathrm{5}}\approx\mathrm{37} \\ $$
Commented by Unhombre last updated on 23/Mar/25
$${yeah}\:{the}\:{answer}\:{is}\:{exactly}\:\mathrm{37} \\ $$$${thanks} \\ $$
Answered by mahdipoor last updated on 23/Mar/25
$${A}=\mathrm{0}.\mathrm{5243}\left(\mathrm{1}+\mathrm{10}^{−\mathrm{4}} +\mathrm{10}^{−\mathrm{8}} +\mathrm{10}^{−\mathrm{12}} +…\right)×\frac{\mathrm{1}−\mathrm{10}^{−\mathrm{4}} }{\mathrm{1}−\mathrm{10}^{−\mathrm{4}} } \\ $$$$\mathrm{0}.\mathrm{5243}\frac{\mathrm{1}−\mathrm{10}^{−\infty} }{\mathrm{1}−\mathrm{10}^{−\mathrm{4}} }\:\:\:\:\left(=\frac{\mathrm{5243}}{\mathrm{9999}}=\right) \\ $$$$\frac{\mathrm{9}}{\mathrm{10}{A}−\mathrm{5}}=\frac{\mathrm{89991}}{\mathrm{2435}}\approx\mathrm{36}.\mathrm{96} \\ $$