{ (( 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.


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Question Number 200315 by Rasheed.Sindhi last updated on 17/Nov/23
${ (( ab ^(−) ∙ b ^(−) + ba ^(−) ∙ a ^(−) = cde ^(−) )),(( ab ^(−) ∙ b ^(−) − ba ^(−) ∙ a ^(−) = f ^(−) )) :} a,b,c,d,e,f are all different and in some order consecutive also. Determine the remaining decimal digits.$
${\begin{cases} \overset{―}{a b} \cdot \overset{―}{b} + \overset{―}{b a} \cdot \overset{―}{a} = \overset{―}{c d e} \\ \overset{―}{a b} \cdot \overset{―}{b} - \overset{―}{b a} \cdot \overset{―}{a} = \overset{―}{f} \end{cases}$ $a, b, c, d, e, f a r e a l l d i f f e r e n t a n d i n$ $s o m e o r d e r c o n s e c u t i v e a l s o .$ $D e t e r m i n e t h e r e m a i n i n g d e c i m a l$ $d i g i t s .$
Commented by Frix last updated on 17/Nov/23

$We can only try . I found$ $a = 3$ $b = 4$ $c = 2$ $d = 6$ $e = 5$ $f = 7$
Commented by nikif99 last updated on 17/Nov/23
$ab^(−) ∙b^− −ba^(−) ∙a^− =f^− ⇒(10a+b)b−(10b+a)a=f ⇒ 10ab+b^2 −10ab−a^2 =f ⇒b^2 −a^2 =f (1) Differences of squares are single digit only for b∈{5,4,3,2} and a∈{4,3,2,1} and b>a (2) ab^(−) ∙b^− +ba^(−) ∙a^− =cde^(−) =(10a+b)b+(10b+a)a= a^2 +b^2 +20ab≥102 (3) According to (2), possible cases for a and b are in the table (blue: a brown: b) determinant ((,1,2,3,4),(2,?,×,×,×),(3,?,∗,×,×),(4,?,□,✓,×),(5,∗,∗,∗,∗)) ×not b>a ? not valid eq. (3) ∗ digit in double occurancy in (3) □ distance of digits >5 in (3) ✓ solution for a=3, b=4, then c=2, d=6,e=5, f=7$
$\overset{―}{a b} \cdot \bar{b} - \overset{―}{b a} \cdot \bar{a} = \bar{f} \Rightarrow (10 a + b) b - (10 b + a) a = f \Rightarrow$ $10 a b + b^{2} - 10 a b - a^{2} = f \Rightarrow b^{2} - a^{2} = f (1)$ $D i f f e r e n c e s o f s q u a r e s a r e s i n g l e$ $d i g i t o n l y f o r b \in {5, 4, 3, 2} a n d a \in {4, 3, 2, 1}$ $a n d b > a (2)$ $\overset{―}{a b} \cdot \bar{b} + \overset{―}{b a} \cdot \bar{a} = \overset{―}{c d e} = (10 a + b) b + (10 b + a) a =$ $a^{2} + b^{2} + 20 a b ⩾ 102 (3)$ $A c c o r d i n g t o (2), p o s s i b l e c a s e s f o r a a n d$ $b a r e i n t h e t a b l e (b l u e : a b r o w n : b)$ $\begin{array}{ccccc} 1 & 2 & 3 & 4 \\ 2 & ? & \times & \times & \times \\ 3 & ? & * & \times & \times \\ 4 & ? & ◻ & ✓ & \times \\ 5 & * & * & * & * \end{array}$ $\times n o t b > a$ $? n o t v a l i d e q . (3)$ $* d i g i t i n d o u b l e o c c u r a n c y i n (3)$ $◻ d i s t a n c e o f d i g i t s > 5 i n (3)$ $✓ s o l u t i o n f o r a = 3, b = 4, t h e n$ $c = 2, d = 6, e = 5, f = 7$
Commented by Rasheed.Sindhi last updated on 18/Nov/23

$\lor \cap i \subset\in!$ $T h a n k s s i r s!$
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