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Question Number 156462 by MathSh last updated on 11/Oct/21

If   abc^(−)  = cba^(−)  + def^(−)   and  a∈{c+2;...;9}  Then find   def^(−)  + fed^(−)

$$\mathrm{If}\:\:\:\overline {\mathrm{abc}}\:=\:\overline {\mathrm{cba}}\:+\:\overline {\mathrm{def}}\:\:\mathrm{and}\:\:\mathrm{a}\in\left\{\mathrm{c}+\mathrm{2};...;\mathrm{9}\right\} \\ $$$$\mathrm{Then}\:\mathrm{find}\:\:\:\overline {\mathrm{def}}\:+\:\overline {\mathrm{fed}}\: \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 11/Oct/21

  let c=1⇒a∈{3,4,...,9}  a=3:  3b1^(−) =1b3^(−) +def^(−)   def^(−) =3b1^(−) −1b3^(−) =301−103=198  def^(−) =198⇒fed^(−) =891   def^(−)  + fed^(−)  =198+891=1089

$$ \\ $$$$\mathrm{let}\:\mathrm{c}=\mathrm{1}\Rightarrow\mathrm{a}\in\left\{\mathrm{3},\mathrm{4},...,\mathrm{9}\right\} \\ $$$$\mathrm{a}=\mathrm{3}: \\ $$$$\overline {\mathrm{3b1}}=\overline {\mathrm{1b3}}+\overline {\mathrm{def}} \\ $$$$\overline {\mathrm{def}}=\overline {\mathrm{3b1}}−\overline {\mathrm{1b3}}=\mathrm{301}−\mathrm{103}=\mathrm{198} \\ $$$$\overline {\mathrm{def}}=\mathrm{198}\Rightarrow\overline {\mathrm{fed}}=\mathrm{891} \\ $$$$\overline {\:\mathrm{def}}\:+\:\overline {\mathrm{fed}}\:=\mathrm{198}+\mathrm{891}=\mathrm{1089} \\ $$

Commented by MathSh last updated on 11/Oct/21

Yes, thank you dear Ser, similary  297→792 ; 396→693 ; 495→594 ...

$$\mathrm{Yes},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{similary} \\ $$$$\mathrm{297}\rightarrow\mathrm{792}\:;\:\mathrm{396}\rightarrow\mathrm{693}\:;\:\mathrm{495}\rightarrow\mathrm{594}\:... \\ $$

Commented by Rasheed.Sindhi last updated on 11/Oct/21

Of course! we can show that, in general  way also.

$$\mathrm{Of}\:\mathrm{course}!\:\mathrm{we}\:\mathrm{can}\:\mathrm{show}\:\mathrm{that},\:\mathrm{in}\:\mathrm{general} \\ $$$$\mathrm{way}\:\mathrm{also}. \\ $$

Commented by MathSh last updated on 11/Oct/21

Yes my dear Ser, thank you so much

$$\mathrm{Yes}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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