Menu Close

ab-b-ba-a-cde-ab-b-ba-a-f-a-b-c-d-e-f-are-all-different-and-in-some-order-consecutive-also-Determine-the-remain




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

Leave a Reply

Your email address will not be published. Required fields are marked *