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Answer - 9.

Yes as per the trial and error method, I got the answer as D. But just be careful in one of the cases...

19 + 92 = 111 cannot be the case that can be suited into our given equation...

We need A B C and D as distinct positive integers...

All others are genuine cases present...

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A has to be 1 as sum of 9+9 = 18 < 20.
Hence B+D = 11
and C+A+1 = 11 => C = 9

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Would this be 700-800 level question?

AB+CD=AAA
A(10)+B+C(10)+D=A(100)+A(10)+A=111A
C10+B+D=101A

At Maximum, C10+B+D must equal less than 90+8+7=105 since all integers are different. Therefore A must equal 1 because it is the only value that will make it less than 105.

If A=1.. Then C must equal 9. If C equaled 8, then 8(10)+9+7=96 does not put it at 101. Nor does anything less than 8. And because C is an single digit integer, then it can't be greater than 10. Therefore, 8<C<10... C=9.

Answer is D.

. AB
+ CD
AAA

if we separately evaluate the left half, A+C must give AA
Logically, take the highest single digit number for ex: 9 =>9+9 = 18 that means, when two sngle numbers are added, the sum can be 18 or less than 18.

so, out of the many possibilities of AA i.e., 11, 22,33 etc, AA can only take the value "11" which is less than 18

in the options, check which value can produce AA = 11.
Ans: 9

(D) 9

A=1

A+C+1=11

A+C=10

C=10-1
C=9

B+D=11
B/D=6/5,7/4,8/3