AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a thre

Question

AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1 
(B) 3
(C) 7
(D) 9
(E) Cannot be determined

Bunuel · Accepted Answer

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 cannot be less than 9, because no two-digit integer with the first digit 1 (1B < 20) can be added to a 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.

Bunuel · Answer

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.

rajathpanta · Answer

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??

Bunuel · Answer

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.

ratinarace · Answer

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?

m3equals333 · Answer

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

PiyushK · Answer

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

GMATinsight · Answer

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.

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

pacifist85 · Answer

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.

mrinal0308 · Answer

Bunuel

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

Bunuel · Answer

Don't know the source but the question is quite GMAT-like.

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

gmatchile1 · Answer

AB + CD = AAA
Since AB and CD are two digit numbers, then AAA must be 111
Therefore 1B + CD = 111
B can assume any value between 3 and 9
If B = 3, then CD = 111-13 = 98 and C = 9
If B = 9, then CD = 111-19 = 92 and C = 9
So for all B between 3 & 9, C = 9
Therefore the correct answer is D (C = 9)

ozzaayy · Answer

It's kinda weird how can we conclude that C will be 9
Only issue is that we need 2 digit numbers with distinct A,B,C, & D
It could be anything like 78,33; 86, 25; and 68, 43

Can someone help me out regarding what's I'm missing here?

Bunuel · Answer

None of your examples meet the condition:

AB + CD = AAA

Notice that A appears as the tens digit in AB and as every digit in the sum, AAA. In light of this, please review the solutions above more carefully.

bumpbot · Answer

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AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a thre

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