2010 AMC 10B Problem 18

Note that $$abc + ab + a = a(bc + b + 1).$$ This means that if $a$ is divisible by $3,$ the whole expression is as well. Since $2010$ is divisible by $3,$ we have that $a$ is divisible by $3$ with probability $\frac{1}{3}.$ Now consider $a$ not divisible by $3.$ For the expression to be divisible by $3,$ we must have that $bc + b + 1$ is divisible by $3.$ This means that $$bc + b = b(c + 1) \equiv 2 \pmod{3}.$$ The only possibility for this is that one of the factors is $2$ mod $3$ and the other is $1$ mod $3.$ For each of the two cases, there is a $\frac{1}{3}$ chance that each of the factors is the desired modulus, for a probability of $$\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}.$$ There are two cases, which means that this happens with a $\dfrac{2}{9}$ probability. The total probability is then $$\dfrac{1}{3} \cdot 1 + \dfrac{2}{3} \cdot \dfrac{2}{9} = \dfrac{13}{27}.$$ Thus, E is the correct answer.