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

{\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

\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!