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#