2) Alberto, Barbara, and Clara are playing in a large square where there are 2008 skittles. Alberto knocks down three times the number of skittles knocked down by Barbara, who in turn knocks down twice the number of skittles knocked down by Clara. What is the maximum number of skittles that Alberto can knock down?
(A) 1321, (B) 1338,
(C) 1342,
(D) 1353,
(E) 1362.
2. The answer is $(\mathbf{B})$.

Let $A, B$ and $C$ be the number of skittles knocked down by Alberto, Barbara, and Clara, respectively. We have $A+B+C \leq 2008$; furthermore, $A=3 B$ and $B=2 C$ so $A=6 C$. Therefore, $6 C+2 C+C=9 C \leq 2008$ from which it follows

$$
C \leq \frac{2008}{9}=223+\frac{1}{9}.
$$

Since $C$ is a natural number, we have that the maximum number of skittles Clara could have knocked down is 223. Consequently, the maximum number of skittles Alberto could have knocked down is $223 \times 6=1338$.