In the long division above, A, B, C, and D represent 4 different...

_______________________________________
The source is GMAT Prep (Focus).

­i did not understand clue 2 and 4 how is 6*d=6?

@gmatophobia 's method is probably the fastest. I just want to add an alternative method incase the long division threw anyone off.


Focusing solely on the top part (ie. \(\frac{ABCB}{6} = BDA\) with a remainder of \(A\)):


From the image one can see that \(6\) does not divide \(a\), and thus \(a < 6\) [Eliminate Answer Choice A]. As \(a\) does not divide by 6, the next step in the division becomes: \(\frac{10a + b}{6} = b\) and becomes \(10a = 5b\) or simplified \(2a = b\). As \(10a + b\) can be written as \(12a\) then \(10a + b\) is divisible by 6, and nothing gets carried over.

Next we have, \(\frac{c}{6}= d\) with a possible remainder. As \(6≤c≤9\), \(d\) can only be \(1\). Let the possible remainder = \(x\). Carry this over to the next step as \(10x\).

Next step becomes: \(\frac{10x + b}{6} = a\) with a remainder of \(a\). However, we know that \(2a = b\):  \(\frac{10x + 2a}{6} = a\) plus the remainder \(a\) becomes: \(10x + 2a = 6a + a\) simplified: \(2x = a\). This shows us that \(a\) is even [Eliminate Answer Choices C & E].

One can rewrite the original division equation as:
­\(\frac{1000a + 100b + 10c + b}{6} = 100b + 10d + a\) with a remainder of \(a\)

Plugging in \(b = 2a\) and \(d = 1\)­:

­\(1000a + 100(2a) + 10c + 2a = 6[100(2a) + 10 + a]+ a\)

­\(1202a + 10c  = 1207a + 60\)

­\(5a + 60 = 10c\)

\(a + 12 = 2c \)

As established from the elimination of answer choices, \(a\) can either be B) \(4\) or D) \(2\). 

If \(a = 4\) , then \(b = 8\) (because of \(2a = b\)) and \(c = 8\). This will not work as the question stem states 'A, B, C, and D represent 4 different integers'

Therefore \(a = 2\)

­ANSWER D
 ­

Am I missing something in this question? It seems fairly straightforward.

ABCB is divisible by 6. A should be less than six. When we multiply B with 6, we should be able to subtract AB without carrying forward a digit as only C is brought forward in the next step.

So, 6 * B = AB which will be a multiple of 6 with 2 digits ~ 12,18,24,30,36,42,48,54,60. We can eliminate 60, 54,42,30,12 as B>A (Divisor has 3 digits only while dividend has more digits for ex 6*600=3600). Now, we are left with 18,24,36,48. Out of the answer choices, we have only 4 and 2 options as for A. So, we eliminate 18,36.

Now, we have 24, 48 for AB values.
Towards the end B - A = A. So, B has to be 4 and A has to be 2?

PS:- You dont have to worry about A,B,C,D (mainly A or B in AB) being 0 as it is not mentioned within the set in the question.­