How to define a, b, c, d, e that are integer between 1-9 so that matrix is a symmetry matrix?

Question

$((4, 10^4a+10^3b+10^2c+10d+e), (4(10^4e+10^3d+10^2c+10b+a), 4))$

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Answer 1 · 2018-04-05T06:01:19

$a = 8, b= 7, c = 9, d = 1, e = 2$

The question essentially is the same as the old puzzle - "Find a five digit number which gets reversed when it is multiplied by 4"

Since there is no carry to the sixth place when "edcba" is multiplied by 4, $e$ must be either 1 or 2.

Again, since $e$ must be $4a$ (modulo 10), the only possibility is $e= 2$ .

Thus $a >= 4e$ - either 8 or9. Of these only 8 leave a remainder of 2 modulo 10, so $a = 8$

This means that $d= 4b+3$ (modulo 10).

On the other hand we have $b >= 4d implies 4d<= 9 implies d<=2$ , but since $d$ must be odd, we have $d=1$

Among 4,5,...,9 , only $b=7$ satisfies $1 = 4b+3$ (modulo 10) and thus $b=7$

Finally, we see that
$4c+3 = 30+c implies c = 9$

So, the numbers are

$a = 8, b= 7, c = 9, d = 1, e = 2$