Since AB and CD are two-digit integers, their sum can give us only one three digit integer of a kind of AAA: 111.

So, A=1 and we have 1B+CD=111

Now, C can not be less than 9, because no two-digit integer with first digit 1 (1B<20) can be added to two-digit integer less than 90, so that to have the sum 111 (if CD<90, so if C<9, CD+1B<111).

Hence C=9.

Answer: D.
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82+19=101, not 111.

As for 111: since AB and CD are two-digit integers, their sum can give us only one three digit integer of a kind of AAA: 111.
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AAA is a 3-digit number with all 3 digits alike, so it could be: 111, 222, 333, ..., 999. But the sum of any two 2-digit numbers cannot be more than 99+99=198, so AB+CD can only be 111.
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\(AB + CD = AAA\)

We can write it down as :

\(A(10) + B + C(10) + D = A(100) + A(10) + A\)

\(B + D + C0 = A00 + A\)

\(B + D + C0 = A0A\)

C0 must be 90 else addition of two single digit B and D wont be able to form a triple digit number e.g if C0=80 then even 80+9+9 = 98.
C = 9
Answer is D

\(B + D + 90 = 101\)

Note : further \(B + D = 11\) and \(A = 1\)
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Well the trick here is to actually realise that AAA emeans that this is the same three digits (in my mind this is not very clear). After you have this straight then even testing does not take long. E.G:

AB + CD = AAA, we can use these numbers for A,B,C,D: 1,2,3,4,5,6,7,8,9

1) Starting with adding the largest numbers possible: 67+89 = we get 156. So we know that the hindreds digit must be one, as we cannot have any larger hundreds digit.

2) This means that AAA = 111. So, A = 1

3) Now we need to surpass 10 in the addition of B and D and also get to 11.

4) Testing shows that 12+ 89 = 101, but 12+78 = 90. So, we should start trying with a combination of 2 low and 2 large numbers.

5) Trying the largest possible for B and D, 9+2 = 11. Visually we have this:
1___B
__+__
C___D
__=__
111

Choose 2 for B and 9 for D (to reach 11) and you will see that it does not work because then you have already used 9 and the second largest number (8) is not enough to reach 11.

Trying the same with 3 for B and 8 for D, you reach 111.

This means that your addition in the end is: 13+98, so C is 9.

AB + CD = AAA.

The tens digit in AB and the digits in AAA are the same. So, A = 1. All this is explained here: https://gmatclub.com/forum/ab-cd-aaa-wh ... l#p1110751
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Don't know the source but the question is quite GMAT-like.

Check other Addition/Subtraction/Multiplication Tables problems from our Special Questions Directory
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Hi..

If it's mentioned that AAA is a 3-digit number and A is a distinct digit, this means that the 3 digits in AAA are all As.
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I think, I can suggest one more way to tackle this problem.
AAA = AB + CD, which implies below:
AAA
- AB
=CD
i.e. 3 digit number - 2 digit number to give 2 digit number.
Hence A<B, because tens digit of AAA and AB are same (if A>=B, we will get 3 digit number as the answer and not 2 digit, which is the case at hand).
Therefore C has to be equal to 9. Also, AAA should be equal to 111 i.e. 111-1B = 9D, where B>1.