If A, B and C are digits between 0 and 9 inclusive, what is

Question

If A, B and C are digits between 0 and 9 inclusive, what is the value of B?

(1) AB + BA = AAC

(2) A = 1.

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Kaplan Math Workbook (6th Edition) official answer states:

statement 1:
that a sum of a two-digit number plus another two-digit number gives a three-digit number. The sum of 2 two-digit numbers must be less that 200, therefore A must be 1.

Then it concludes that this statement is true only when b=9 ----> Sufficient

Statement 2 is Insufficient (obviously) and therefore the AC is A.

Evaluating the red phrase: it assumes that A and B must be two-digit numbers. Isn't it possible that A=0? Isn't the following addition correct? (B has to equal 0 for it to be correct, but still...)

+ 0 B
+ B 0
-------
0 0 B

Or because no one writes the 0's therefore it implies that A and B can't equal 0?

I would really appreciate any comments.

Show Answer

Official Answer

A

Bunuel · Accepted Answer

I think one thing we can assume safely is that A, B and C are distinct digits.

Next, even if we don't assume that A>0 (because of AB and AAC as you mention) then still A can not be zero as 0B+B0=00C doesn't hold true for any values of B and C (single digit number B plus two-digit number B0 can not equal to C, also a single digit number).

If A, B and C are digits between 0 and 9 inclusive, what is the value of B?

(1) AB + BA = AAC --> well from above A=1 (A can not be more than 1, as the sum of 2 two-digit numbers is always less than 200) --> 1B+B1=11C --> B=9 (B can not be less than 8 as 18+81=99, not a three digit number) --> 19+91=110. Sufficient.

(2) A = 1. Not sufficient.

Answer: A.

Hope it helps.

SonyGmat · Answer

Thanks for your post Bunuel! +1

why should we assume that they are distinct digits?

Let me try to illustrate my point with another example:

Can the sum of two prime numbers, A and B, be an even number?

Of course it can. If a=b=2 then 2+2=4. It doesn't state that x and y are distinct values, therefore to my understanding, I can assume that they could have the same value.

Now back to my post:

Since the question stem doesn't state that A,B,C are distinct values, why should we assume it? isn't it possible for A=B=C=0?

Bunuel · Answer

I agree that it would have been better if they mentioned "distinct" in the stem (to avoid confusion in some test-takers). Having said that, in order your case to be valid 1. we shouldn't consider digits to be distinct (which was quite obvious at least for me) and 2. consider 00+00=000 as a proper option, which is too "technical" and non-conventional way of writing digit 0.

Anyway since you understand what was the intended meaning of the question I wouldn't worry about these technicalities at all. On the GMAT you won't see any ambiguous questions at all.

dentobizz · Answer

I have a question about statement number:2]--A=1

If A equals one then B has to be 9 , any other number <9, will not give a carryover figure that is required to get AA or (11)

so I think statement:2 should also be sufficient. Am I missing something here?

Bunuel · Answer

You can not use information given in statement (1) to for statement (2). So, the only thing we know from (2) is that A=1, which is clearly insufficient to answer the question.

Hope it's clear.

dentobizz · Answer

yes it is clear, thank you. I thought the equation given was part of the question.

manjunath0128 · Answer

The best way to approach this problem is,
1. A and B are interchanged 2 digit number.
2. Most important is, 2 statements cannot contradict each other. Hence A should be equal to 1( Though it is possible that the equation satisfies 1 and any other number). If you plug the number in original equation A= 1 and B =9. Thus A is sufficient.
Also, since A and B are 2 digit number, we can use
10A+10B + 10B + A ===>
11(A+B).
Right hand side must be multiple of 11 and 3 digit number.

MS26 · Answer

Given A,B&C are all single digit 0 to 9 now going the statement
1. Given AB+BA=AAC , we can write as (10A+B)+(10B+A)=(100A+10A+C), 11A+11B=110A+C, 11B=99A+C, B=9A+(C/11), since B is given between 0-9 so only way this is possible is when A=1 and C=0 so  A is sufficient ,
2. A=1  alone   not sufficient

If A, B and C are digits between 0 and 9 inclusive, what is

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If A, B and C are digits between 0 and 9 inclusive, what is

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