2010 AMC 10B Problem 18

Whether a number is a multiple of $3$ depends only on the remainders when you divide by $3$. So think about remainders.

1. Factor. Every term has an $a$, so
$$abc + ab + a = a(bc + b + 1).$$
A product is a multiple of $3$ exactly when one of its factors is (because $3$ is prime).
2. Each remainder is equally likely. $2010 = 3 \cdot 670$, so among $1, \dots, 2010$ exactly $670$ numbers leave remainder $0$, $670$ leave remainder $1$, and $670$ leave remainder $2$. Each remainder has probability $\frac13$, separately for $a$, $b$ and $c$.
3. Case 1: $a$ is a multiple of $3$. Probability $\frac13$. The whole expression is then a multiple of $3$.
4. Case 2: $a$ is not a multiple of $3$. Probability $\frac23$. Now we need $bc + b + 1$ to be a multiple of $3$. Check all $9$ pairs of remainders for $b$ and $c$ (only remainders matter):

Two of the $9$ equally likely pairs work, so this has probability $\frac29$.
5. Add the two cases (they do not overlap):
$$\frac13 + \frac23 \cdot \frac29 = \frac{9}{27} + \frac{4}{27} = \frac{13}{27}.$$

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

Divisibility by $3$ only cares about remainders, so a problem about $2010$ choices becomes a problem about $3$ remainders. Factoring out the common $a$ splits the question into "is $a$ a multiple of $3$?" and "is the other factor?". The number $2010$ matters only because it is a multiple of $3$, which makes all three remainders equally likely.

Count remainder triples. There are $3 \cdot 3 \cdot 3 = 27$ equally likely choices of remainders for $(a, b, c)$.

• $a$ has remainder $0$: any $b$ and $c$ work, $9$ triples.
• $a$ has remainder $1$ or $2$: only the $2$ good pairs from the table work, $2 \cdot 2 = 4$ triples.

That is $9 + 4 = 13$ good triples out of $27$, so the probability is $\frac{13}{27}$.