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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Hi Bunuel,

Doesn't 82+19=111. But you say no two digit number with 1st digit as 1 can be added to a number less than 90 to get a sum of 111. Can you elaborate??

One more thing is only 111 fits into this scenario as it is the sum of two digit numbers which cannot exceed 188 right??

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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Hi Bunuel

How did u check that there is only one such number as AAA when AB and CD are added together..Please can you explain the logic behind it?

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

10A+B+10C+D=100A+10A+A
10C+B+D=101A
(10C+B+D)/101=A

Just need to realize that A,B,C,D are integers that cannot exceed 9, as they are single digits

Keeping this in mind, C is at most 9, so (90+B+D)/101=A -->B+D=11 and A=1

For A = 2, it becomes obvious that either B, C, or D will have to exceed 9 and is thus not possible.

So C=9
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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 \(A = 1\)

Answer in Two steps

Step 1: AB + CD = AAA i.e. Sum of two 2-digit numbers is a three digit number but sum of two 2-digit number must be less than 200 hence the 3-Digit number must have Hundreds digit as 1 i.e. A = 1

Step 2: If one 2-Digit number is in 10s i.e. less than 20 (AB here as A=1) then other two digit number must be in either 80s or 90s in order to give the sum of two greater than 100 i.e. CD must have C either 8 or 9 only. [Since number D in CD is unknown so it gives us opportunity to make summation 111 anyways]

Of 8 and 9, only 9 is the available option
Answer: Option
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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.

Please ignore

One thing i am clear that the no has to be 111. But why is 24+87=111 wrong?

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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Bunuel

Is this a GMAT like question?
What is the actual source of this question?

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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chetan2u

Please help.

The question says that AAA is a 3-digit number. Nowhere is it mentioned that AAA have the same digits. How it is possible to narrow down AAA to 111?

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.

A really interesting question, Even we can make 4 cases such as

AB
1D

AB
3D

AB
7D

AB
9D

If we look at case 4, only this can satisfy our condition, A, B, C, and D are distinct positive integers

_16
+95
111
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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.

AB
CD
------
AAA
------

look carefully at B+D: it gives 10 + A. So B+D=10 + A

Now lets get into the calculation part:

10A + B +10C + D = 100A + 10A + A

10A + 10 + A + 10C = 111A

10 + 10 C = 100A

10(1+C) = 100A

1+ C = 10A

It is given that a,b,c and d are single digit nos. So in order for that statement to hold true. The value of A can only be 1.

So, 1 + c = 10

C= 9 .option D is the correct answer.


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