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

## $\left(\begin{matrix}4 & {10}^{4} a + {10}^{3} b + {10}^{2} c + 10 d + e \\ 4 \left({10}^{4e+10} ^ 3 d + {10}^{2} c + 10 b + a\right) & 4\end{matrix}\right)$

Apr 5, 2018

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

#### Explanation:

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

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

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

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

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

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

So, the numbers are

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