Question Number 200315 by Rasheed.Sindhi last updated on 17/Nov/23
$$\:\begin{cases}{\overline {\:{ab}\:}\centerdot\overline {\:{b}\:}+\overline {\:{ba}\:}\centerdot\overline {\:{a}\:}=\overline {\:{cde}\:}}\\{\overline {\:{ab}\:}\centerdot\overline {\:{b}\:}−\overline {\:{ba}\:}\centerdot\overline {\:{a}\:}=\overline {\:{f}\:}\:}\end{cases} \\ $$$${a},{b},{c},{d},{e},{f}\:{are}\:{all}\:{different}\:{and}\:{in} \\ $$$${some}\:{order}\:{consecutive}\:{also}. \\ $$$$\: \\ $$$$\mathcal{D}{etermine}\:{the}\:{remaining}\:{decimal} \\ $$$${digits}. \\ $$
Commented by Frix last updated on 17/Nov/23
$$\mathrm{We}\:\mathrm{can}\:\mathrm{only}\:\mathrm{try}.\:\mathrm{I}\:\mathrm{found} \\ $$$${a}=\mathrm{3} \\ $$$${b}=\mathrm{4} \\ $$$${c}=\mathrm{2} \\ $$$${d}=\mathrm{6} \\ $$$${e}=\mathrm{5} \\ $$$${f}=\mathrm{7} \\ $$
Commented by nikif99 last updated on 17/Nov/23
$$\overline {{ab}}\centerdot\overset{−} {{b}}−\overline {{ba}}\centerdot\overset{−} {{a}}=\overset{−} {{f}}\:\Rightarrow\left(\mathrm{10}{a}+{b}\right){b}−\left(\mathrm{10}{b}+{a}\right){a}={f}\:\Rightarrow \\ $$$$\mathrm{10}{ab}+{b}^{\mathrm{2}} −\mathrm{10}{ab}−{a}^{\mathrm{2}} ={f}\:\Rightarrow{b}^{\mathrm{2}} −{a}^{\mathrm{2}} ={f}\:\left(\mathrm{1}\right) \\ $$$${Differences}\:{of}\:{squares}\:{are}\:{single} \\ $$$${digit}\:{only}\:{for}\:{b}\in\left\{\mathrm{5},\mathrm{4},\mathrm{3},\mathrm{2}\right\}\:{and}\:{a}\in\left\{\mathrm{4},\mathrm{3},\mathrm{2},\mathrm{1}\right\} \\ $$$${and}\:{b}>{a}\:\left(\mathrm{2}\right) \\ $$$$\overline {{ab}}\centerdot\overset{−} {{b}}+\overline {{ba}}\centerdot\overset{−} {{a}}=\overline {{cde}}=\left(\mathrm{10}{a}+{b}\right){b}+\left(\mathrm{10}{b}+{a}\right){a}= \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{20}{ab}\geqslant\mathrm{102}\:\left(\mathrm{3}\right) \\ $$$${According}\:{to}\:\left(\mathrm{2}\right),\:{possible}\:{cases}\:{for}\:{a}\:{and}\: \\ $$$${b}\:{are}\:{in}\:{the}\:{table}\:\left({blue}:\:{a}\:{brown}:\:{b}\right) \\ $$$$\begin{array}{|c|c|c|c|c|}{}&\hline{\mathrm{1}}&\hline{\mathrm{2}}&\hline{\mathrm{3}}&\hline{\mathrm{4}}\\{\mathrm{2}}&\hline{?}&\hline{×}&\hline{×}&\hline{×}\\{\mathrm{3}}&\hline{?}&\hline{\ast}&\hline{×}&\hline{×}\\{\mathrm{4}}&\hline{?}&\hline{\square}&\hline{\checkmark}&\hline{×}\\{\mathrm{5}}&\hline{\ast}&\hline{\ast}&\hline{\ast}&\hline{\ast}\\\hline\end{array} \\ $$$$×{not}\:{b}>{a} \\ $$$$?\:{not}\:{valid}\:{eq}.\:\left(\mathrm{3}\right) \\ $$$$\ast\:{digit}\:{in}\:{double}\:{occurancy}\:{in}\:\left(\mathrm{3}\right) \\ $$$$\square\:{distance}\:{of}\:{digits}\:>\mathrm{5}\:{in}\:\left(\mathrm{3}\right) \\ $$$$\checkmark\:{solution}\:{for}\:{a}=\mathrm{3},\:{b}=\mathrm{4},\:{then} \\ $$$${c}=\mathrm{2},\:{d}=\mathrm{6},{e}=\mathrm{5},\:{f}=\mathrm{7} \\ $$
Commented by Rasheed.Sindhi last updated on 18/Nov/23
$$\vee\:\cap\boldsymbol{\mathrm{i}}\subset\in! \\ $$$$\mathcal{T}{hanks}\:\boldsymbol{{sirs}}! \\ $$