solution to last week's MGRE Math Beast Challenge

MGRE has published its solution to last week's quantitative comparison problem. My answer of C was correct. Here's MGRE's discussion of the solution, which differs from mine:

Since x, y, and z are consecutive multiples of 3, let’s write all three integers in terms of x:
x = x
y = x + 3
z = x + 6

So, for Quantity A, the sum can be rewritten only in terms of x:
(x + 1) + (y – 2) + (z + 3)
(x + 1) + (x + 3 – 2) + (x + 6 + 3) {substituting for y and z}
3x + (1 + 3 – 2 + 6 + 3)
3x + 11

Now we find the remainder when this simplified sum is divided by 9.
(Remainder when (3x + 11) div 9) = (Remainder when 3x div 9) + (Remainder when 11 div 9)

Since x is a multiple of 3, 3x is a multiple of 9, and the remainder when 3x is divided by 9 is 0.
When 11 is divided by 9, 9 goes into 11 once, leaving 11 – 9 = 2 as the remainder.
Thus, Quantity A is 0 + 2 = 2.

The correct answer is C.

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