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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