Question Number 10521 by ajfour last updated on 16/Feb/17
$${A}\:{number}\:\left(\alpha\beta..\lambda…\mu\mathrm{2}\right)×\mathrm{2}\:=\left(\mathrm{2}\alpha\beta..\lambda…\mu\right) \\ $$$${find}\:{the}\:{number}. \\ $$$$ \\ $$
Answered by mrW1 last updated on 17/Feb/17
$$\left({Part}\:{I}\right) \\ $$$${let}\:{x}=\left(\alpha\beta..\lambda…\mu\right)={a}\:{number}\:{with}\:{n}−\mathrm{1}\:{digits} \\ $$$${let}\:{y}=\left(\alpha\beta..\lambda…\mu\mathrm{2}\right)={a}\:{number}\:{with}\:{n}\:{digits} \\ $$$${let}\:{z}=\left(\mathrm{2}\alpha\beta..\lambda…\mu\right)={a}\:{number}\:{with}\:{n}\:{digits} \\ $$$${y}={x}×\mathrm{10}+\mathrm{2}=\mathrm{10}{x}+\mathrm{2} \\ $$$${z}=\mathrm{2}×\mathrm{10}…\mathrm{0}+{x}=\mathrm{2}×\mathrm{10}^{{n}−\mathrm{1}} +{x} \\ $$$$ \\ $$$$\left(\alpha\beta..\lambda…\mu\mathrm{2}\right)×\mathrm{2}\:=\left(\mathrm{2}\alpha\beta..\lambda…\mu\right) \\ $$$$\Leftrightarrow{y}×\mathrm{2}={z} \\ $$$$\left(\mathrm{10}{x}+\mathrm{2}\right)×\mathrm{2}=\mathrm{2}×\mathrm{10}^{{n}−\mathrm{1}} +{x} \\ $$$$\mathrm{19}{x}=\mathrm{2}×\mathrm{10}^{{n}−\mathrm{1}} −\mathrm{4}=\mathrm{2}\left(\mathrm{10}^{{n}−\mathrm{1}} −\mathrm{2}\right) \\ $$$${x}=\frac{\mathrm{2}\left(\mathrm{10}^{{n}−\mathrm{1}} −\mathrm{2}\right)}{\mathrm{19}} \\ $$$$ \\ $$$${the}\:{first}\:{suitable}\:{number}\:{is}\:{n}=\mathrm{18}\:{and} \\ $$$${x}=\frac{\mathrm{2}\left(\mathrm{10}^{\mathrm{17}} −\mathrm{2}\right)}{\mathrm{19}}=\mathrm{10526315789473684} \\ $$$${y}=\mathrm{105263157894736842} \\ $$$$\left(\mathrm{105263157894736842}\right)×\mathrm{2}=\left(\mathrm{210526315789473684}\right) \\ $$$$ \\ $$$${the}\:{second}\:{suitable}\:{number}\:{is}\:{n}=\mathrm{36}\:{and} \\ $$$${x}=\frac{\mathrm{2}\left(\mathrm{10}^{\mathrm{35}} −\mathrm{2}\right)}{\mathrm{19}}=\mathrm{10526315789473684210526315789473684} \\ $$$${y}=\mathrm{105263157894736842105263157894736842} \\ $$$$\left(\mathrm{105263157894736842105263157894736842}\right)×\mathrm{2}=\left(\mathrm{210526315789473684210526315789473684}\right) \\ $$$$ \\ $$$${the}\:{third}\:{suitable}\:{number}\:{is}\:{n}=\mathrm{54}\:{and} \\ $$$${x}=\frac{\mathrm{2}\left(\mathrm{10}^{\mathrm{53}} −\mathrm{2}\right)}{\mathrm{19}}=\mathrm{10526315789473684210526315789473684210526315789474684} \\ $$$${y}=\mathrm{105263157894736842105263157894736842105263157894736842} \\ $$$$\left(\mathrm{105263157894736842105263157894736842105263157894736842}\right)×\mathrm{2}=\left(\mathrm{210526315789473684210526315789473684210526315789473684}\right) \\ $$$$ \\ $$$${the}\:{general}\:{solution}\:{is}\:{n}=\mathrm{18}{i}\:{with}\:{i}=\mathrm{1},\mathrm{2},… \\ $$$${and}\:{the}\:{number}\:{is} \\ $$$${y}=\mathrm{105263157894736842105263157894736842105263157894736842105263157894736842}…… \\ $$
Commented by ajfour last updated on 16/Feb/17
$${great}\:{explanation},\:{i}\:{cud}\:{arrive}\:{at}\:{the}\:{answer}\:{but}\:{ur}\:{explanation}\:{is}\:{elaborate}.\:{but}\:{wait},\:{it}\:{should}\:{end}\:{with}….\:\mathrm{736842}.{typing}\:{error}\:{probably}. \\ $$
Commented by mrW1 last updated on 16/Feb/17
$${you}\:{are}\:{right}.\:{i}\:{had}\:{a}\:{typing}\:{error}.\: \\ $$$${now}\:{it}'{s}\:{fixed}.\:{thank}\:{you}\:{for}\:{pointing}\:{out}\:{that}. \\ $$