2010 AMC 10B Problem 18

Factor: $abc + ab + a = a(bc + b + 1) = a\big(b(c+1) + 1\big)$.

Because $2010 = 3 \cdot 670$, each of $a, b, c$ is equally likely to be $0, 1$, or $2 \pmod 3$, independently.

The product is divisible by $3$ if $a \equiv 0$, which has probability $\frac13$.

Otherwise (probability $\frac23$) we need $b(c+1) + 1 \equiv 0$, i.e. $b(c+1) \equiv 2 \pmod 3$. The residue pairs $(b, c+1)$ with product $2$ are $(1, 2)$ and $(2, 1)$; each pair occurs with probability $\frac19$, so this has probability $\frac29$.

Total:
$$\frac13 + \frac23\cdot\frac29 = \frac{9}{27} + \frac{4}{27} = \frac{13}{27}.$$

The answer is $\boxed{\textbf{(E)}\ \dfrac{13}{27}}$.

Divisibility of an expression depends only on residues, and factoring exposes the structure: a product is $0 \bmod 3$ when some factor is. The size $2010$ matters only through being a multiple of $3$, which makes the three residues uniform; then the problem is a small tree of cases on residues.

Complement: the expression is not divisible by $3$ when $a \not\equiv 0$ (probability $\frac23$) and $b(c+1) \not\equiv 2$ (probability $\frac79$), giving $\frac{14}{27}$ of failure and $1 - \frac{14}{27} = \frac{13}{27}$ of success.

• Forgetting the residues are uniform only because 2010 is a multiple of 3, or double counting the cases where a is divisible by 3.
• Treating $bc + b + 1 \equiv 0$ as requiring $b \equiv 0$ or $c \equiv 0$, which is backwards: if $b \equiv 0$ the bracket is $1$.