Hi Bunuel VeritasKarishma

Can you please provide solution to the above problem?
Sum of the number and its reverse is a multiple of 11 but here all the numbers are divisible by 11. So I believe this concept doesn't apply here. How can we solve it?

Let 'abcdef' is a 6-digit number.
sum of a 6-digit number and the number formed by reversing its digits= 100000(a+f)+10000(b+e)+1000(c+d)+100(c+d)+10(b+e)+(a+f)

Hence we can notice that difference between Hundred-thousands digit and unit digit can be 0 (if b+e<10), or 1 (if b+e>9).

In option D difference between Hundred-thousands digit and unit digit is 9-7=2
hence 951467 can never be the sum of a 6-digit number and the number formed by reversing its digits.

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). Option (A) works if (a+f) = 7, option (B) works if (a+f) = 12, option (C) works if (a+f) = 3 (and there is a carry over of 1 from b+e). Option (E) works if (a+f) = 9.
Option (D) does not work since if (a+f) = 7, the first digit cannot be 9. Even with a carry over, it can only be 8. Two single digits can never add up to give a carry over of 2.

Answer (D)
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