Let the numbers be abcdef and fedcba. When we add them we get
100000(a+f) + 10000(b+e) + 1000(c+d) + 100(d+c) +10(b+e) + (f+a)
= 100001(a+f) + 10010(b+e) + 1100(c+d)
Whatever (a+f) is, that will be the first digit and the last digit. (a+f) can be anything from 2 (=1+1) to 18 (=9+9)
Moreover,Add the numbers, club them, and you’ll see that the sum is divisible by 11.
Only option D satisfies this
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