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Question Number 10521 by ajfour last updated on 16/Feb/17

A number (αβ..λ...μ2)×2 =(2αβ..λ...μ) find the number.

$${A}\:{number}\:\left(\alpha\beta..\lambda...\mu\mathrm{2}\right)×\mathrm{2}\:=\left(\mathrm{2}\alpha\beta..\lambda...\mu\right) \\ $$$${find}\:{the}\:{number}. \\ $$$$ \\ $$

Answered by mrW1 last updated on 17/Feb/17

(Part I) let x=(αβ..λ...μ)=a number with n−1 digits let y=(αβ..λ...μ2)=a number with n digits let z=(2αβ..λ...μ)=a number with n digits y=x×10+2=10x+2 z=2×10...0+x=2×10^(n−1) +x (αβ..λ...μ2)×2 =(2αβ..λ...μ) ⇔y×2=z (10x+2)×2=2×10^(n−1) +x 19x=2×10^(n−1) −4=2(10^(n−1) −2) x=((2(10^(n−1) −2))/(19)) the first suitable number is n=18 and x=((2(10^(17) −2))/(19))=10526315789473684 y=105263157894736842 (105263157894736842)×2=(210526315789473684) the second suitable number is n=36 and x=((2(10^(35) −2))/(19))=10526315789473684210526315789473684 y=105263157894736842105263157894736842 (105263157894736842105263157894736842)×2=(210526315789473684210526315789473684) the third suitable number is n=54 and x=((2(10^(53) −2))/(19))=10526315789473684210526315789473684210526315789474684 y=105263157894736842105263157894736842105263157894736842 (105263157894736842105263157894736842105263157894736842)×2=(210526315789473684210526315789473684210526315789473684) the general solution is n=18i with i=1,2,... and the number is y=105263157894736842105263157894736842105263157894736842105263157894736842......

$$\left({Part}\:{I}\right) \\ $$$${let}\:{x}=\left(\alpha\beta..\lambda...\mu\right)={a}\:{number}\:{with}\:{n}−\mathrm{1}\:{digits} \\ $$$${let}\:{y}=\left(\alpha\beta..\lambda...\mu\mathrm{2}\right)={a}\:{number}\:{with}\:{n}\:{digits} \\ $$$${let}\:{z}=\left(\mathrm{2}\alpha\beta..\lambda...\mu\right)={a}\:{number}\:{with}\:{n}\:{digits} \\ $$$${y}={x}×\mathrm{10}+\mathrm{2}=\mathrm{10}{x}+\mathrm{2} \\ $$$${z}=\mathrm{2}×\mathrm{10}...\mathrm{0}+{x}=\mathrm{2}×\mathrm{10}^{{n}−\mathrm{1}} +{x} \\ $$$$ \\ $$$$\left(\alpha\beta..\lambda...\mu\mathrm{2}\right)×\mathrm{2}\:=\left(\mathrm{2}\alpha\beta..\lambda...\mu\right) \\ $$$$\Leftrightarrow{y}×\mathrm{2}={z} \\ $$$$\left(\mathrm{10}{x}+\mathrm{2}\right)×\mathrm{2}=\mathrm{2}×\mathrm{10}^{{n}−\mathrm{1}} +{x} \\ $$$$\mathrm{19}{x}=\mathrm{2}×\mathrm{10}^{{n}−\mathrm{1}} −\mathrm{4}=\mathrm{2}\left(\mathrm{10}^{{n}−\mathrm{1}} −\mathrm{2}\right) \\ $$$${x}=\frac{\mathrm{2}\left(\mathrm{10}^{{n}−\mathrm{1}} −\mathrm{2}\right)}{\mathrm{19}} \\ $$$$ \\ $$$${the}\:{first}\:{suitable}\:{number}\:{is}\:{n}=\mathrm{18}\:{and} \\ $$$${x}=\frac{\mathrm{2}\left(\mathrm{10}^{\mathrm{17}} −\mathrm{2}\right)}{\mathrm{19}}=\mathrm{10526315789473684} \\ $$$${y}=\mathrm{105263157894736842} \\ $$$$\left(\mathrm{105263157894736842}\right)×\mathrm{2}=\left(\mathrm{210526315789473684}\right) \\ $$$$ \\ $$$${the}\:{second}\:{suitable}\:{number}\:{is}\:{n}=\mathrm{36}\:{and} \\ $$$${x}=\frac{\mathrm{2}\left(\mathrm{10}^{\mathrm{35}} −\mathrm{2}\right)}{\mathrm{19}}=\mathrm{10526315789473684210526315789473684} \\ $$$${y}=\mathrm{105263157894736842105263157894736842} \\ $$$$\left(\mathrm{105263157894736842105263157894736842}\right)×\mathrm{2}=\left(\mathrm{210526315789473684210526315789473684}\right) \\ $$$$ \\ $$$${the}\:{third}\:{suitable}\:{number}\:{is}\:{n}=\mathrm{54}\:{and} \\ $$$${x}=\frac{\mathrm{2}\left(\mathrm{10}^{\mathrm{53}} −\mathrm{2}\right)}{\mathrm{19}}=\mathrm{10526315789473684210526315789473684210526315789474684} \\ $$$${y}=\mathrm{105263157894736842105263157894736842105263157894736842} \\ $$$$\left(\mathrm{105263157894736842105263157894736842105263157894736842}\right)×\mathrm{2}=\left(\mathrm{210526315789473684210526315789473684210526315789473684}\right) \\ $$$$ \\ $$$${the}\:{general}\:{solution}\:{is}\:{n}=\mathrm{18}{i}\:{with}\:{i}=\mathrm{1},\mathrm{2},... \\ $$$${and}\:{the}\:{number}\:{is} \\ $$$${y}=\mathrm{105263157894736842105263157894736842105263157894736842105263157894736842}...... \\ $$

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.... 736842.typing error probably.

$${great}\:{explanation},\:{i}\:{cud}\:{arrive}\:{at}\:{the}\:{answer}\:{but}\:{ur}\:{explanation}\:{is}\:{elaborate}.\:{but}\:{wait},\:{it}\:{should}\:{end}\:{with}....\:\mathrm{736842}.{typing}\:{error}\:{probably}. \\ $$

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.

$${you}\:{are}\:{right}.\:{i}\:{had}\:{a}\:{typing}\:{error}.\: \\ $$$${now}\:{it}'{s}\:{fixed}.\:{thank}\:{you}\:{for}\:{pointing}\:{out}\:{that}. \\ $$

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